Cut a region into thin strips of area $y\,dx$, add them up and you get $\displaystyle\int_a^b y\,dx$. Turn the strips sideways for curves of the form $x=g(y)$, and take the modulus wherever the curve dips below the axis โ an area is never negative.
A drone flies along a straight road while its sensor records the height of a stretch of embankment above the road every few metres. An engineer needs the area of the embankment's cross-section to estimate how much soil it contains. Or a farmer needs the area of a field whose fourth side follows a curving canal. In both cases the boundary is a curve $y=f(x)$, and the plan is the same: cut the region into thin vertical strips, find the area of one strip, and add. In Chapter 7 โ Integrals you met the idea that such a sum, in the limit of infinitely thin strips, is the definite integral. This section turns that idea into a reliable method for finding areas โ and shows you the traps the method hides.
There are two of them. First, some regions are much easier to cut into horizontal strips, because their boundary is naturally written as $x=g(y)$ โ the parabola $y^2=4ax$ is the standard example. Second, the integral of a function that lies below the $x$-axis is negative, and the integral of one that crosses the axis can be zero, even though the region plainly has a positive area. Board examiners know this trap well; after this section you will never fall into it.
Elementary Strips and the Area under a Curve
Suppose $f$ is continuous and $f(x)\ge0$ for $a\le x\le b$, and let $R$ be the region bounded by the curve $y=f(x)$, the $x$-axis and the vertical lines $x=a$ and $x=b$. Picture a very thin vertical strip of $R$ standing on the $x$-axis at position $x$, with width $dx$. Its top edge is a tiny piece of the curve, but because the strip is so thin the top is almost level, at height $y=f(x)$. So the strip is almost a rectangle, and its area is $$dA=y\,dx.$$ The area of $R$ is the total of all these strip areas as $x$ runs from $a$ to $b$ โ and "adding up $y\,dx$ from $a$ to $b$" is exactly what $\displaystyle\int_a^b y\,dx$ means.
๐ Definition 8.1 โ Region under a curve; elementary strip
Let $f$ be continuous on $[a,b]$ with $f(x)\ge0$ there. The region under the curve $y=f(x)$ from $x=a$ to $x=b$ is the set of points $(x,y)$ with $a\le x\le b$ and $0\le y\le f(x)$: it is bounded by the curve, the $x$-axis and the lines $x=a$, $x=b$. Its area is called the area under the curve.
An elementary strip of the region is a thin rectangle used to build its area. A vertical strip at $x$ has width $dx$ and height $y=f(x)$, so area $dA=y\,dx$. A horizontal strip at height $y$ has thickness $dy$ and length $x$, so area $dA=x\,dy$.
Why does the sum of strips give exactly the integral, and not just an approximation? Let $A(x)$ be the area under the curve from $a$ up to a moving right-hand edge at $x$, so that $A(a)=0$ and the area we want is $A(b)$. Push the edge from $x$ to $x+h$ (with $h\gt0$ small). The extra area $A(x+h)-A(x)$ is one strip of width $h$; it lies between a rectangle of height $m$ and one of height $M$, where $m$ and $M$ are the least and greatest values of $f$ on $[x,x+h]$: $$m\,h\le A(x+h)-A(x)\le M\,h,$$ so, dividing by $h$, $$m\le\dfrac{A(x+h)-A(x)}{h}\le M.$$ As $h\to0$ both $m$ and $M$ tend to $f(x)$, because $f$ is continuous. (The same works for $h\lt0$.) Hence $A'(x)=f(x)$: the area function is an antiderivative of $f$. If $F$ is any antiderivative, $A(x)=F(x)+C$, and $A(a)=0$ gives $C=-F(a)$. So the area is $A(b)=F(b)-F(a)=\displaystyle\int_a^b f(x)\,dx$.
๐ Theorem 8.1 โ Area under a curve (vertical strips)
If $f$ is continuous on $[a,b]$ and $f(x)\ge0$ for all $x$ in $[a,b]$, then the area of the region bounded by the curve $y=f(x)$, the $x$-axis and the lines $x=a$ and $x=b$ is $$A=\int_a^b y\,dx=\int_a^b f(x)\,dx=F(b)-F(a),$$ where $F$ is any antiderivative of $f$. Areas are measured in square units (mยฒ, cmยฒ, โฆ) when $x$ and $y$ are lengths.
โ๏ธ Example 1 โ Area under a straight line
Find the area of the region bounded by the line $y=2x+1$, the $x$-axis and the lines $x=1$ and $x=3$. Check your answer with the formula for the area of a trapezium.
๐ก Show solutionโถ
Sketch For $1\le x\le3$, $y=2x+1$ runs from $3$ to $7$, so the line is above the $x$-axis and Theorem 8.1 applies. A vertical strip at $x$ has area $dA=(2x+1)\,dx$.
Check The region is a trapezium with parallel vertical sides $3$ and $7$ and width $2$: area $\tfrac12(3+7)\times2=10$ โ.
โ๏ธ Example 2 โ Areas under parabolas
(a) Find the area bounded by $y=x^2$, the $x$-axis and the lines $x=-1$ and $x=2$. (b) Find the area of the region in the first quadrant bounded by the parabola $y^2=4x$, the $x$-axis and the lines $x=1$ and $x=4$.
๐ก Show solutionโถ
(a) $x^2\ge0$ everywhere, so $$A=\int_{-1}^2x^2\,dx=\Big[\dfrac{x^3}{3}\Big]_{-1}^2=\dfrac83-\Big(-\dfrac13\Big)=3\ \text{square units}.$$ (The piece from $-1$ to $0$ contributes $\tfrac13$ and the piece from $0$ to $2$ contributes $\tfrac83$; both are positive.)
(b) the right branch $y^2=4x$ is a parabola opening to the right. In the first quadrant $y\ge0$, so the upper half is $y=2\sqrt x$ โ a function of $x$, which vertical strips need.
Check At $x=1$ the height is $2$ and at $x=4$ it is $4$; the region sits inside the $3\times4$ rectangle (area $12$) and contains the $3\times2$ one (area $6$). $9.33$ lies between โ.
Horizontal Strips: Area between a Curve and the $y$-axis
Look at the region bounded by the parabola $x=y^2$, the $y$-axis and the line $y=2$. Vertical strips are awkward here: each one runs from the curve up to the line $y=2$, not down to the $x$-axis. Turn the strips sideways instead. A horizontal strip at height $y$, of thickness $dy$, runs from the $y$-axis to the curve, so its length is $x=y^2$ and its area is $dA=x\,dy$. Exactly the same argument as before, with the roles of $x$ and $y$ swapped, gives the next theorem.
๐ Theorem 8.2 โ Area between a curve and the $y$-axis (horizontal strips)
If $g$ is continuous on $[c,d]$ and $g(y)\ge0$ for all $y$ in $[c,d]$, then the area of the region bounded by the curve $x=g(y)$, the $y$-axis and the lines $y=c$ and $y=d$ is $$A=\int_c^d x\,dy=\int_c^d g(y)\,dy.$$
Use it whenever the boundary is given (or easily written) as $x=g(y)$ โ for example $y^2=4ax$, $x=4-y^2$, or $y=x^2$ in the first quadrant, which is $x=\sqrt y$.
โ๏ธ Example 3 โ Choosing horizontal strips
(a) Find the area of the region in the first quadrant bounded by $y=x^2$, the $y$-axis and the line $y=4$. Then check it with vertical strips. (b) Find the area bounded by the curve $x=y^2+1$, the $y$-axis and the lines $y=0$ and $y=3$.
๐ก Show solutionโถ
(a) set up The region lies to the left of the curve, between the $y$-axis and the parabola, for $0\le y\le4$. In the first quadrant $y=x^2$ means $x=\sqrt y$, so a horizontal strip has area $dA=\sqrt y\,dy$.
(a) check With vertical strips: the region is the $2\times4$ rectangle $0\le x\le2$, $0\le y\le4$ with the area under $y=x^2$ removed. That is $8-\displaystyle\int_0^2x^2\,dx=8-\tfrac83=\tfrac{16}{3}$ โ. Notice that $\tfrac{16}{3}$ and $\tfrac83$ split the rectangle in the ratio $2:1$.
(b) $g(y)=y^2+1\ge1\gt0$, so the curve is to the right of the $y$-axis: $$A=\int_0^3(y^2+1)\,dy=\Big[\dfrac{y^3}{3}+y\Big]_0^3=9+3=12\ \text{square units}.$$
Interactive 1 โ Sweep a Strip, Watch the Area Grow
Choose a region. The red strip sweeps across it โ vertical strips $dA=y\,dx$ or horizontal strips $dA=x\,dy$ โ and everything it has passed is shaded blue: that is the running area $A$. Compare the strip's height (or length) with the rate at which $A$ is growing: they are the same number, which is why the area is the integral.
strip positionโ
strip height / lengthโ
rate of growth of $A$โ
running area $A$โ
whole regionโ
Curves below the $x$-axis: an Area Is Never Negative
Now let the curve dip below the axis. Take $y=x^2-4x$ for $0\le x\le4$: since $x^2-4x=x(x-4)$, it is negative between $0$ and $4$. A vertical strip there still has a positive area, but its "height" $y$ is negative, so $y\,dx$ is negative: the integral counts the region with a minus sign. Indeed $\displaystyle\int_0^4(x^2-4x)\,dx=\tfrac{64}{3}-32=-\tfrac{32}{3}$. The area is the size of that number, $\tfrac{32}{3}$. In general the definite integral measures the net area โ area above the axis minus area below it โ while an area question asks for the total, with every part counted positively.
๐ Theorem 8.3 โ Curves below or crossing the $x$-axis
Let $f$ be continuous on $[a,b]$.
(i) If $f(x)\le0$ for all $x$ in $[a,b]$, the area bounded by $y=f(x)$, the $x$-axis and the lines $x=a$, $x=b$ is $$A=-\int_a^b f(x)\,dx=\left|\int_a^b f(x)\,dx\right|.$$
(ii) If $f$ changes sign only at the points $c_1\lt c_2\lt\dots\lt c_k$ inside $(a,b)$, split $[a,b]$ at these points. On each piece $f$ keeps one sign, so $$A=\left|\int_a^{c_1}f(x)\,dx\right|+\left|\int_{c_1}^{c_2}f(x)\,dx\right|+\dots+\left|\int_{c_k}^{b}f(x)\,dx\right|=\int_a^b|f(x)|\,dx.$$
The same holds for horizontal strips: a curve $x=g(y)$ to the left of the $y$-axis gives a negative $\displaystyle\int_c^d g(y)\,dy$, and its area is the modulus.
โ๏ธ Example 4 โ A region entirely below the axis
Find the area of the region bounded by the parabola $y=x^2-4$ and the $x$-axis.
๐ก Show solutionโถ
Sketch $x^2-4=0$ at $x=\pm2$. The parabola opens upwards with vertex $(0,-4)$, so between $-2$ and $2$ it lies below the axis. The bounded region is the "bowl" between the curve and the axis for $-2\le x\le2$.
Answer By Theorem 8.3(i), $A=\left|-\dfrac{32}{3}\right|=\dfrac{32}{3}\approx10.67$ square units. Writing "area $=-\tfrac{32}{3}$" loses the mark.
Check The region fits inside the $4\times4$ rectangle $-2\le x\le2$, $-4\le y\le0$ (area $16$), and $\tfrac{32}{3}$ is exactly $\tfrac23$ of $16$ โ the parabola fact Archimedes discovered, which you will meet again in Section 8.2.
โ๏ธ Example 5 โ When the integral is zero but the area is not
(a) Evaluate $\displaystyle\int_0^{2\pi}\sin x\,dx$, and find the area bounded by $y=\sin x$ and the $x$-axis from $x=0$ to $x=2\pi$. (b) Find the area bounded by $y=x^3$, the $x$-axis and the lines $x=-1$ and $x=2$.
๐ก Show solutionโถ
(a) the integral $\displaystyle\int_0^{2\pi}\sin x\,dx$ $=\Big[-\cos x\Big]_0^{2\pi}$ $=-1-(-1)=0$.
(a) the area $\sin x$ is positive on $(0,\pi)$ and negative on $(\pi,2\pi)$: split at $\pi$. $\displaystyle\int_0^{\pi}\sin x\,dx=\Big[-\cos x\Big]_0^{\pi}=1+1=2$ and $\displaystyle\int_\pi^{2\pi}\sin x\,dx=-1-1=-2$. So $A=|2|+|-2|=4$ square units. The two arches have equal areas and cancel in the integral โ but not in the area.
(b) $x^3$ changes sign only at $0$. $\displaystyle\int_{-1}^0x^3\,dx=\Big[\dfrac{x^4}4\Big]_{-1}^0=-\dfrac14$ and $\displaystyle\int_0^2x^3\,dx=\dfrac{16}{4}=4$. So $A=\dfrac14+4=\dfrac{17}{4}$ square units.
Compare $\displaystyle\int_{-1}^2x^3\,dx=4-\tfrac14=\tfrac{15}{4}$ โ the net area. It is smaller than the area $\tfrac{17}{4}$ by twice the area below the axis.
โ๏ธ Example 6 โ Distance travelled from a velocity graph
A lift moves along a vertical shaft with velocity $v(t)=t^2-4t+3$ metres per second (upwards positive) for $0\le t\le3$ seconds. Find (a) its displacement and (b) the total distance it travels in these $3$ seconds. Interpret both as statements about areas.
๐ก Show solutionโถ
Idea On the velocityโtime graph a thin strip of width $dt$ has "area" $v\,dt$, the small displacement in time $dt$. The displacement is the integral (net area); the distance counts every metre travelled, up or down โ the area.
(a) displacement $\displaystyle\int_0^3(t^2-4t+3)\,dt$ $=\Big[\dfrac{t^3}{3}-2t^2+3t\Big]_0^3$ $=9-18+9=0\ \text{m}$. The lift ends where it started.
(b) distance $v(t)=(t-1)(t-3)$ is positive for $0\lt t\lt1$ and negative for $1\lt t\lt3$. $\displaystyle\int_0^1v\,dt=\tfrac13-2+3=\tfrac43$ and $\displaystyle\int_1^3v\,dt=0-\tfrac43=-\tfrac43$. Distance $=\tfrac43+\tfrac43=\tfrac83\approx2.67\ \text{m}$.
Interpret The lift rises $\tfrac43$ m during the first second, stops at $t=1$, and comes back down $\tfrac43$ m. The area above the $t$-axis equals the area below it.
Interactive 2 โ The Sign Trap: Net Integral versus Area
Choose a curve and move the right-hand end $b$ (or press Play). Parts above the axis are shaded blue and counted $+$ by the integral; parts below are shaded red and counted $-$. The area adds both as positive amounts. Watch for the moments when the integral is $0$ while the area is not.
intervalโ
roots insideโ
net integral $\int_a^b f\,dx$โ
area $\int_a^b |f|\,dx$โ
๐งญ Problem-solving strategy โ finding an area with integrals
Sketch the curve and the bounding lines, and shade the region. Mark where the curve meets the axes.
Choose the strip. Vertical ($dA=y\,dx$) if the region runs from the $x$-axis up (or down) to a curve $y=f(x)$; horizontal ($dA=x\,dy$) if it runs from the $y$-axis across to a curve $x=g(y)$.
Find the limits โ the extreme positions of the strip โ from the bounding lines or from where the curve meets the axis.
Check the sign of $y$ (or $x$) on the interval. If it changes, split at the crossing points (Theorem 8.3).
Use symmetry if the region is symmetric about an axis: find one part and multiply.
Integrate, take the modulus of every negative piece, add, and write square units. Sanity-check against a rectangle that contains the region.
โ ๏ธ Common mistakes when studying alone
1. Integrating straight across a root. $\displaystyle\int_0^{2\pi}\sin x\,dx=0$, but the area is $4$. Always find where the curve crosses the axis before integrating.
2. Taking the modulus of the whole integral. $\left|\displaystyle\int_{-1}^2x^3\,dx\right|=\tfrac{15}4$ is still wrong; the area is $\tfrac{17}4$. The modulus goes on each piece, not on the total.
3. A negative final answer. An area can never be negative. If you get $-\tfrac{32}3$, you integrated over a part below the axis: the area is $\tfrac{32}3$.
4. Using the whole parabola $y^2=4x$ as if it were $y=f(x)$. For vertical strips you need $y$ as a function of $x$: the upper half is $y=2\sqrt x$, the lower half $y=-2\sqrt x$. Or use horizontal strips with $x=\tfrac{y^2}{4}$.
5. Mixing up the variables. With horizontal strips the limits are $y$-values and the integrand must be written in terms of $y$. $\displaystyle\int_0^4\sqrt x\,dy$ is meaningless.
6. Forgetting units. If $x$ and $y$ are in metres, the area is in mยฒ. Write "square units" when no units are given.
Exercises
I
Vertical strips. Find the area of the region bounded by the given curve, the $x$-axis and the given lines (sketch first).
1 $y=3x$; $x=0$, $x=4$
2 $y=x+2$; $x=-2$, $x=2$ (check with the area of a triangle)
3 $y=x^2$; $x=1$, $x=3$
4 $y=9-x^2$, between the points where it meets the $x$-axis
5 $y^2=9x$ in the first quadrant; $x=0$, $x=4$
6 $y=\sqrt{x-1}$; $x=1$, $x=5$
7 $y=x^3$; $x=0$, $x=2$
8 $y=\cos x$; $x=0$, $x=\tfrac\pi2$
๐ก Show solutionsโถ
1 $\displaystyle\int_0^4 3x\,dx=\Big[\tfrac32x^2\Big]_0^4=24$ square units (a triangle with base $4$ and height $12$ โ).
2 $x+2\ge0$ on $[-2,2]$: $\displaystyle\int_{-2}^2(x+2)\,dx=\Big[\tfrac{x^2}2+2x\Big]_{-2}^2=6-(-2)=8$. Triangle: base $4$, height $4$, area $8$ โ.
4 $9-x^2=0$ at $x=\pm3$, and the curve is above the axis between them. $\displaystyle\int_{-3}^3(9-x^2)\,dx$ $=\Big[9x-\tfrac{x^3}3\Big]_{-3}^3$ $=18-(-18)=36$ square units.
8 $\cos x\ge0$ on $[0,\tfrac\pi2]$: $\Big[\sin x\Big]_0^{\pi/2}=1$ square unit.
II
Horizontal strips. Find the area of the region bounded by the given curve, the $y$-axis and the given lines.
1 $x=y^2$; $y=1$, $y=3$
2 $y=x^2$ in the first quadrant; $y=1$, $y=4$
3 $y=x^3$ in the first quadrant; $y=1$, $y=8$
4 $x=4-y^2$ (the whole region between the curve and the $y$-axis)
5 the line $x+2y=8$; $y=0$, $y=2$ (check with the area of a trapezium)
6 Find the area of the region bounded by $y=\sqrt x$, the $y$-axis and the line $y=2$ in two ways: with horizontal strips, and as a rectangle minus the area under a curve.
๐ก Show solutionsโถ
1 $\displaystyle\int_1^3y^2\,dy=\Big[\tfrac{y^3}3\Big]_1^3=9-\tfrac13=\tfrac{26}3$ square units (the same number as I.3 โ the region is I.3's region reflected in $y=x$).
4 $4-y^2\ge0$ for $-2\le y\le2$: $\displaystyle\int_{-2}^2(4-y^2)\,dy$ $=\Big[4y-\tfrac{y^3}3\Big]_{-2}^2$ $=\tfrac{16}3-\big(-\tfrac{16}3\big)=\tfrac{32}3$ square units.
5 $x=8-2y$: $\displaystyle\int_0^2(8-2y)\,dy=\Big[8y-y^2\Big]_0^2=12$. Trapezium with parallel sides $8$ (at $y=0$) and $4$ (at $y=2$), height $2$: $\tfrac12(8+4)\cdot2=12$ โ.
6 Horizontal: $x=y^2$, $\displaystyle\int_0^2y^2\,dy=\tfrac83$. Rectangle: the region is the rectangle $0\le x\le4$, $0\le y\le2$ minus the area under $y=\sqrt x$: $8-\displaystyle\int_0^4\sqrt x\,dx=8-\tfrac{16}3=\tfrac83$ โ square units.
III
Curves below or crossing the axis. Find the area bounded by the curve, the $x$-axis and the given lines (and the net integral where asked).
1 $y=x^2-9$, between its $x$-intercepts
2 $y=-\sqrt x$; $x=0$, $x=4$
3 $y=x-3$; $x=0$, $x=5$ (check with two triangles)
4 $y=\cos x$; $x=0$, $x=\pi$. Also give $\displaystyle\int_0^\pi\cos x\,dx$.
5 $y=x^2-x-2$; $x=0$, $x=3$. Also give the net integral.
6 $y=x^3-4x$; $x=-2$, $x=2$
7 True or false? Give a reason. "If $\displaystyle\int_a^b f(x)\,dx=0$, then the area bounded by $y=f(x)$, the $x$-axis and $x=a$, $x=b$ is $0$."
๐ก Show solutionsโถ
1 Intercepts $\pm3$; the curve is below the axis between them. $\displaystyle\int_{-3}^3(x^2-9)\,dx$ $=\Big[\tfrac{x^3}3-9x\Big]_{-3}^3$ $=-18-18=-36$, so the area is $36$ square units.
2 $\displaystyle\int_0^4-\sqrt x\,dx=-\tfrac{16}3$; area $\tfrac{16}3$ square units (the reflection in the $x$-axis of the region under $y=\sqrt x$).
3 Root at $x=3$. $\displaystyle\int_0^3(x-3)\,dx=\Big[\tfrac{x^2}2-3x\Big]_0^3=-\tfrac92$ and $\displaystyle\int_3^5(x-3)\,dx=\Big(\tfrac{25}2-15\Big)-\Big(-\tfrac92\Big)=2$. Area $=\tfrac92+2=\tfrac{13}2$ square units. Triangles: $\tfrac12\cdot3\cdot3+\tfrac12\cdot2\cdot2=\tfrac92+2$ โ.
4 $\displaystyle\int_0^\pi\cos x\,dx=\Big[\sin x\Big]_0^\pi=0$. Split at $\tfrac\pi2$: $\displaystyle\int_0^{\pi/2}\cos x\,dx=1$, $\displaystyle\int_{\pi/2}^\pi\cos x\,dx=-1$. Area $=2$ square units.
5 $x^2-x-2=(x-2)(x+1)$: negative on $(0,2)$, positive on $(2,3)$. With $F(x)=\tfrac{x^3}3-\tfrac{x^2}2-2x$: $F(0)=0$, $F(2)=-\tfrac{10}3$, $F(3)=-\tfrac32$. Pieces: $-\tfrac{10}3$ and $-\tfrac32+\tfrac{10}3=\tfrac{11}6$. Area $=\tfrac{10}3+\tfrac{11}6=\tfrac{31}6$ square units; net integral $=-\tfrac32$.
6 $x^3-4x=x(x-2)(x+2)$: positive on $(-2,0)$, negative on $(0,2)$. $\displaystyle\int_{-2}^0(x^3-4x)\,dx$ $=\Big[\tfrac{x^4}4-2x^2\Big]_{-2}^0$ $=0-(4-8)=4$, and by the odd symmetry the piece on $[0,2]$ is $-4$. Area $=8$ square units (net integral $0$).
7 False. $\displaystyle\int_0^{2\pi}\sin x\,dx=0$, but the area is $4$ (Example 5). A zero integral only says the areas above and below the axis are equal.
IV
Applications and unknown limits.
1Braking cyclist. A cyclist's velocity along a straight road is $v(t)=6-2t$ m/s for $0\le t\le5$ s (she brakes, stops and then rolls back down a slope). Find her displacement and the total distance travelled.
2Canal-side field. A field is bounded by a straight road (the $x$-axis), two walls $x=0$ and $x=40$, and a canal bank $y=30+0.5x-0.01x^2$ (all in metres). Find its area.
3Filling a tank. Water flows into a tank at $r(t)=2+0.3t$ litres per minute. Interpret $\displaystyle\int_0^{10}r(t)\,dt$ as an area and find how much water enters in the first $10$ minutes.
4 Find $k\gt0$ if the area bounded by $y=x^2$, the $x$-axis and the lines $x=0$, $x=k$ is $9$ square units.
5 Find $a\gt0$ if the area bounded by $y=\sqrt x$, the $x$-axis and the lines $x=0$, $x=a$ is $18$ square units.
6 The line $x=c$ divides the region under $y=x^2$ from $x=0$ to $x=3$ into two parts of equal area. Find $c$.
7 Find the area bounded by $y=|x|$, the $x$-axis and the lines $x=-1$ and $x=3$.
๐ก Show solutionsโถ
1 Displacement $\displaystyle\int_0^5(6-2t)\,dt=\Big[6t-t^2\Big]_0^5=5$ m. $v=0$ at $t=3$: $\displaystyle\int_0^3=9$ and $\displaystyle\int_3^5=5-9=-4$. Distance $=9+4=13$ m.
2 The bank stays above the road ($y=30$ at $x=0$, $34$ at $x=40$, and $y\gt0$ between). $\displaystyle\int_0^{40}(30+0.5x-0.01x^2)\,dx$ $=\Big[30x+0.25x^2-\tfrac{0.01x^3}{3}\Big]_0^{40}$ $=1200+400-213.33$ $\approx1386.7\ \text{m}^2$ (exactly $\tfrac{4160}{3}$ mยฒ).
3 A strip of width $dt$ under the rate graph has area $r\,dt$ = litres entering in time $dt$. $\displaystyle\int_0^{10}(2+0.3t)\,dt$ $=\Big[2t+0.15t^2\Big]_0^{10}$ $=20+15=35$ litres.
4 $\tfrac{k^3}3=9$, so $k^3=27$ and $k=3$.
5 $\tfrac23a^{3/2}=18$, so $a^{3/2}=27$ and $a=27^{2/3}=9$.
6 Whole area $\tfrac{27}3=9$; we need $\tfrac{c^3}3=\tfrac92$, so $c^3=\tfrac{27}2$ and $c=\dfrac{3}{\sqrt[3]2}\approx2.381$. (Not $1.5$: the region is much taller on the right.)
7 $|x|\ge0$ everywhere; split at $0$ where the formula changes: $\displaystyle\int_{-1}^0(-x)\,dx+\int_0^3x\,dx=\tfrac12+\tfrac92=5$ square units (two triangles).
How this is examined: Board papers ask for the area bounded by a simple curve, an axis and given lines โ usually with a sketch, a correct choice of strip, the right limits and a positive answer in square units; MCQs test the sign trap. JEE builds on the same set-up with unknown limits and parameters.
E1.Board ยท 1 mark The area of the region bounded by $y=x^3$, the $x$-axis and the lines $x=-1$ and $x=1$ is (A) $0$ (B) $\tfrac14$ (C) $\tfrac12$ (D) $1$
๐ก Show solutionโถ
Idea $x^3$ is negative on $(-1,0)$ and positive on $(0,1)$; split at $0$. Each piece has modulus $\tfrac14$.
Answer (C) $\tfrac14+\tfrac14=\tfrac12$. Trap (A) is $\displaystyle\int_{-1}^1x^3\,dx$, the net area; (B) is only one half of the region.
E2.Board ยท 2 marks Find the area of the region in the first quadrant bounded by the parabola $x^2=4y$, the $y$-axis and the line $y=3$.
๐ก Show solutionโถ
Set up The region lies between the $y$-axis and the curve for $0\le y\le3$: horizontal strips, with $x=2\sqrt y$ in the first quadrant. $A=\displaystyle\int_0^3 2\sqrt y\,dy$. (1 mark)
E3.Board ยท 3 marks Sketch the curve $y=x^2-2x$ and find the area bounded by it, the $x$-axis and the lines $x=0$ and $x=3$.
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Sketch $y=x(x-2)$: an upward parabola through $(0,0)$ and $(2,0)$ with vertex $(1,-1)$. It is below the axis on $(0,2)$ and above it on $(2,3)$. (1 mark)
Split With $F(x)=\tfrac{x^3}3-x^2$: $\displaystyle\int_0^2=F(2)-F(0)=\tfrac83-4=-\tfrac43$ and $\displaystyle\int_2^3=F(3)-F(2)=0-\big(-\tfrac43\big)=\tfrac43$. (1 mark)
Answer $A=\tfrac43+\tfrac43=\tfrac83$ square units. (1 mark) Notice that $\displaystyle\int_0^3(x^2-2x)\,dx=0$: a student who does not split gets $0$ and loses all three marks.
E4.Board ยท Case study The entrance of a pedestrian underpass has the shape of the region under the arch $y=4-\tfrac{x^2}{4}$ above the road $y=0$ (lengths in metres, $x=0$ at the centre). (i) Find where the arch meets the road and the area of the entrance. [2 marks] (ii) Use symmetry to write the area of the right half. [1 mark] (iii) A glass panel fills the part of the entrance between $x=2$ and the right-hand edge of the arch. Find its area. [1 mark]
๐ก Show solutionโถ
(i) $4-\tfrac{x^2}4=0$ at $x=\pm4$. $A=\displaystyle\int_{-4}^4\Big(4-\tfrac{x^2}4\Big)dx$ $=\Big[4x-\tfrac{x^3}{12}\Big]_{-4}^4$ $=2\Big(16-\tfrac{64}{12}\Big)$ $=\tfrac{64}3\approx21.3\ \text{m}^2$.
(ii) The arch is symmetric about $x=0$ (the $y$-axis), so the right half has area $\tfrac{32}3\approx10.7\ \text{m}^2$.
E5.JEE Main The area of the region $\{(x,y):0\le y\le x^2+1,\ 0\le x\le2\}$ is (A) $\tfrac83$ (B) $\tfrac{11}3$ (C) $\tfrac{14}3$ (D) $\tfrac{16}3$
๐ก Show solutionโถ
Method Read the set: $x$ runs from $0$ to $2$, and for each $x$ the point goes from the $x$-axis up to the curve $y=x^2+1\gt0$. That is the region under the curve: $\displaystyle\int_0^2(x^2+1)\,dx=\tfrac83+2$.
Answer (C) $\tfrac{14}3$. Trap (A) forgets the "$+1$" (the $2\times1$ rectangle underneath the parabola). โฑ Fast route: the area under $x^2$ on $[0,2]$ is $\tfrac13$ of the $2\times4$ box, i.e. $\tfrac83$; add the rectangle $2$.
E6.JEE Advanced(Numerical answer.) For $k\gt0$, the curve $y=x^2-kx$ and the $x$-axis bound a region of area $36$ square units. Find $k$.
๐ก Show solutionโถ
Set up $x^2-kx=x(x-k)$ meets the axis at $0$ and $k$ and is below it in between. $\displaystyle\int_0^k(x^2-kx)\,dx=\tfrac{k^3}3-\tfrac{k^3}2=-\tfrac{k^3}6$, so the area is $\tfrac{k^3}6$.
Answer $\tfrac{k^3}6=36$ gives $k^3=216$, so $k=\mathbf{6}$. โฑ Fast route: the region is $\tfrac23$ of the bounding box of width $k$ and depth $\tfrac{k^2}4$ (the vertex is at $x=\tfrac k2$), so $\tfrac23\cdot\tfrac{k^3}4=\tfrac{k^3}6$.
๐ฌ ResearchโTeaching NexusMini project ยท about 60 minutes
Research question. How accurately can you measure the area of a leaf with strips โ and does a parabola model of its edge give the same answer as counting squares?
Why it matters. Leaf area controls how much sunlight a plant captures and how much water it loses. Agricultural scientists use leaf-area measurements to estimate crop growth and irrigation needs, and ecologists compare leaf areas across climates. Professional leaf-area meters do exactly what this section does: they scan the leaf in thin strips and add up the widths.
What to do.
Trace a leaf (mango, peepal, neemโฆ) on 1 mm graph paper with its midrib along the $x$-axis. Count whole and half squares to get a reference area.
At every $1$ cm along the midrib, measure the width $w(x)$ of the leaf (across both halves). Enter $x$ and $w$ in a spreadsheet.
Estimate the area as the sum of strip areas $\sum w\,\Delta x$ with $\Delta x=1$ cm. Repeat with $\Delta x=2$ cm (every other reading). Which estimate is closer to the square count?
Model the leaf: if its length is $L$ and greatest width $W$, try $w(x)=\dfrac{4W}{L^2}x(L-x)$ and show that $\displaystyle\int_0^L w(x)\,dx=\tfrac23LW$. Compare this model with your measurements.
Tools: Graph paper, ruler, a few leaves, spreadsheet (or a phone photo with a 1 cm grid behind the leaf).
Hand in: Your table of widths, the two strip estimates, the square count, the $\tfrac23LW$ model value, the percentage differences, and two sentences on which leaves the parabola model suits and which it does not.
Going further. Measure ten leaves from the same tree and plot area against $L\times W$. Is the ratio close to $\tfrac23$? Botanists use such ratios to estimate leaf area from two quick measurements.
Instructor note: for a leaf with $L=15$ cm and $W=5$ cm the parabola model gives $\tfrac23\cdot75=50$ cmยฒ. Strip sums with $1$ cm spacing usually agree with the square count to within $2$โ$5\%$; real leaves are often slightly fuller than a parabola near the base, giving ratios between about $0.65$ and $0.72$.
๐งช Self-check quiz
0/5 correct
Q1. For the region under $y=f(x)\ge0$, the area of a thin vertical strip at position $x$ is
Q2. The area bounded by $y=x^2$, the $x$-axis and the lines $x=0$ and $x=3$ is
Q3. The area bounded by $y=\sin x$ and the $x$-axis from $x=0$ to $x=2\pi$ is
Q4. The area bounded by $y=-x^3$, the $x$-axis and the lines $x=0$ and $x=2$ is
Q5. If $g$ is continuous and $g(y)\ge0$ on $[c,d]$, the area bounded by $x=g(y)$, the $y$-axis and the lines $y=c$, $y=d$ is