7.4 Definite Integrals, the Fundamental Theorem and Their Properties
The definite integral measures the net area under a curve. The fundamental theorem of calculus says the rate at which that area grows is the height of the curve โ so every area is just $F(b)-F(a)$ for an antiderivative $F$. Seven properties then make many hard-looking integrals easy.
A train's speed during a $10$-minute run between two Mumbai suburban stations is recorded on a graph. How far apart are the stations? For a constant speed you would multiply speed by time โ the area of a rectangle under the speed graph. For a changing speed, the distance is still the area under the speedโtime graph, but now the top edge is curved. The same question appears everywhere: the total rainfall from a rainfall-rate chart, the energy used from a power meter's graph, the area of an irregular field bounded by a curved river bank.
So far, integration has meant โfinding antiderivativesโ. This section introduces a second, apparently quite different idea โ the definite integral $\displaystyle\int_a^bf(x)\,dx$, the net area between a curve and the $x$-axis โ and then shows, in the fundamental theorem of calculus, that the two ideas are one. The area under $f$ from $a$ to $b$ is $F(b)-F(a)$ for any antiderivative $F$. Everything from Sections 7.1โ7.3 therefore becomes a way of computing areas, and Chapter 8 will use it to find the areas of regions bounded by parabolas, circles and ellipses.
The Definite Integral as an Area
๐ Definition 7.5 โ Definite integral
Let $f$ be continuous on a closed interval $[a,b]$. The definite integral of $f$ from $a$ to $b$, written
$$\int_a^bf(x)\,dx$$
is the net (signed) area of the region between the graph $y=f(x)$, the $x$-axis and the lines $x=a$, $x=b$: parts of the region above the $x$-axis count as positive, parts below as negative. The numbers $a$ and $b$ are the lower limit and upper limit of integration, and $[a,b]$ is the interval of integration.
Unlike $\int f(x)\,dx$, which is a family of functions, $\int_a^bf(x)\,dx$ is a number. How can we find that number when the top edge is curved? One way is to fill the region with thin rectangles, as in Interactive 1: each rectangle has an area we can compute, and as the rectangles get thinner the total settles down to a single value โ the definite integral. (Turning this into a formal definition is a topic beyond the Board syllabus; see the ๐ box at the end of this section.) For some regions, ordinary geometry already gives the answer.
โ๏ธ Example 1 โ Definite integrals from geometry
Use areas of familiar shapes to find (a) $\displaystyle\int_0^32x\,dx$ (b) $\displaystyle\int_{-2}^{2}\sqrt{4-x^2}\,dx$ (c) $\displaystyle\int_0^{2\pi}\sin x\,dx$.
๐ก Show solutionโถ
(a) The region under $y=2x$ from $0$ to $3$ is a triangle with base $3$ and height $6$: area $\tfrac12\cdot3\cdot6=9$.
(b) $y=\sqrt{4-x^2}$ is the upper half of the circle $x^2+y^2=4$. The region is a semicircle of radius $2$: area $\tfrac12\pi\cdot2^2=2\pi$.
(c) The arch above the axis on $[0,\pi]$ is the mirror image of the arch below it on $[\pi,2\pi]$. Their signed areas cancel: the integral is $0$, even though the total shaded area is not zero.
Lesson A definite integral is a net area. When you want the actual area of a region, split the interval where the curve crosses the axis and add the absolute values (Chapter 8).
Interactive 1 โ Thin Rectangles Settle on the Definite Integral
Choose a function and a rule for the height of each rectangle: its left end, right end or midpoint. Rectangles above the axis are blue (positive), below are red (negative). Increase $n$, or press Play to double it again and again: the total of the rectangles approaches one number, the definite integral. The last readout gives that number exactly, from the fundamental theorem below.
$n$ rectangles, widthโ
total of rectanglesโ
exact: $F(b)-F(a)$โ
The Area Function and the Fundamental Theorems
Instead of one fixed right-hand end $b$, let the right-hand end move. Then the area becomes a function of where we stop.
๐ Definition 7.6 โ Area function
Let $f$ be continuous on $[a,b]$. For each $x$ in $[a,b]$, the area function of $f$ from $a$ is
$$A(x)=\int_a^xf(t)\,dt$$
the net area under $f$ from $a$ up to $x$. (The variable inside is called $t$ so that it is not confused with the moving end $x$.) Note that $A(a)=0$.
๐ Theorem 7.11 โ First fundamental theorem of calculus
If $f$ is continuous on $[a,b]$, then the area function $A(x)=\displaystyle\int_a^xf(t)\,dt$ is differentiable on $(a,b)$ and
$$A'(x)=f(x)\quad\text{for every }x\in(a,b).$$
In words: the rate at which the area grows, as the right-hand end moves, is the height of the curve at that end. So $A$ is an antiderivative of $f$.
Why (a picture argument). Move the right-hand end from $x$ to $x+h$ for a small $h\gt0$. The area grows by the area of a thin strip of width $h$ under the curve:
$$A(x+h)-A(x)=\int_x^{x+h}f(t)\,dt$$
Because $f$ is continuous, on that narrow strip $f(t)$ stays close to $f(x)$, so the strip is almost a rectangle of height $f(x)$ and width $h$: $A(x+h)-A(x)\approx f(x)\,h$. Dividing by $h$, $\dfrac{A(x+h)-A(x)}{h}\approx f(x)$, and the approximation becomes exact as $h\to0$. That limit is $A'(x)$. Watch the thin strip and the slope of $A$ in Interactive 2.
๐ Theorem 7.12 โ Second fundamental theorem of calculus
Let $f$ be continuous on $[a,b]$ and let $F$ be any antiderivative of $f$ on $[a,b]$. Then
$$\int_a^bf(x)\,dx=F(b)-F(a)$$
The right-hand side is written $\big[F(x)\big]_a^b$.
Why. By Theorem 7.11, $A$ is an antiderivative of $f$; so is $F$. By Theorem 7.1 they differ by a constant: $A(x)=F(x)+C$. Putting $x=a$: $0=A(a)=F(a)+C$, so $C=-F(a)$ and $A(x)=F(x)-F(a)$. Putting $x=b$ gives $\displaystyle\int_a^bf(x)\,dx=A(b)=F(b)-F(a)$.
Two practical consequences. First, any antiderivative works, and the constant of integration cancels, so we leave it out: $\big[F(x)+C\big]_a^b=F(b)-F(a)$. Second, the theorem needs $f$ to be continuous on the whole of $[a,b]$ โ check this before you use it.
Interactive 2 โ The Area Function Grows at the Rate f(x)
Left: $f$, with the net area from $a$ to $x$ shaded (blue above the axis, red below) and the next thin strip in purple. Right: the area function $A(x)$ being drawn, with its tangent line at $x$. Play moves $x$. Notice that $A$ rises where $f\gt0$, falls where $f\lt0$, and is flat exactly where $f=0$: the slope of $A$ is always the height of $f$. That is the first fundamental theorem.
$f(t)$ and the area from $a$ to $x$
$A(x)=\int_a^xf(t)\,dt$ and its slope
$x$โ
$A(x)$ (area so far)โ
slope of $A$ and $f(x)$โ
โ๏ธ Example 2 โ Evaluating with the second fundamental theorem
(c) $\big[\tan^{-1}x\big]_0^1=\tfrac\pi4-0=\tfrac\pi4$ โ so the area under $\dfrac{1}{1+x^2}$ from $0$ to $1$ is exactly $\tfrac\pi4\approx0.7854$. Compare Interactive 1.
(d) $\big[\log x\big]_1^e=1-0=1$.
Check (a) by Interactive 1's method: the midpoint rule with $n=4$ on $x^2+1$ gives $3.328\ldots$, close to $\tfrac{10}3=3.333\ldots$.
โ๏ธ Example 3 โ Using the methods of Sections 7.1โ7.3
(a) By parts, an antiderivative is $(x-1)e^x$. So $\big[(x-1)e^x\big]_0^1=0-(-1)=1$.
(b) By special integral (1) with $a=1$:
$$\left[\frac12\log\left|\frac{x-1}{x+1}\right|\right]_2^3=\frac12\left(\log\frac12-\log\frac13\right)=\frac12\log\frac32$$
The integrand is continuous on $[2,3]$ (its only breaks are at $\pm1$), so the theorem applies.
Check (c) $\approx0.1427$. Since $0\le\sin^2x\le\tfrac12$ on $[0,\tfrac\pi4]$, the answer must lie between $0$ and $\tfrac\pi8\approx0.39$ โ.
Substitution in Definite Integrals
With a substitution $u=g(x)$ you could find an antiderivative in $u$, put $x$ back and then use the limits. It is quicker to change the limits as well, and never go back to $x$ at all.
๐ Theorem 7.13 โ Substitution with limits
Let $g$ have a continuous derivative on $[a,b]$ and let $f$ be continuous on an interval containing all the values $g(x)$, $a\le x\le b$. Then
$$\int_a^bf(g(x))\,g'(x)\,dx=\int_{g(a)}^{g(b)}f(u)\,du$$
Why. If $F'=f$, then $F(g(x))$ is an antiderivative of the left integrand (Theorem 7.4). By Theorem 7.12 both sides equal $F(g(b))-F(g(a))$. In practice: when $x=a$, $u=g(a)$; when $x=b$, $u=g(b)$ โ write the new limits next to the substitution.
โ๏ธ Example 4 โ Change the limits, not just the variable
(b) $u=\cos x$, $du=-\sin x\,dx$; $x=0\Rightarrow u=1$, $x=\tfrac\pi2\Rightarrow u=0$:
$$\int_1^0\frac{-du}{1+u^2}=\int_0^1\frac{du}{1+u^2}=\frac\pi4$$
(Swapping the limits absorbs the minus sign โ property Pโ below.)
(c) $u=x+2$, so $x=u-2$, $dx=du$; the limits become $2$ and $4$:
$$\int_2^4(u-2)u^{1/2}\,du=\left[\tfrac25u^{5/2}-\tfrac43u^{3/2}\right]_2^4$$
At $u=4$: $\tfrac{64}5-\tfrac{32}3=\tfrac{32}{15}$. At $u=2$: $\tfrac{8\sqrt2}5-\tfrac{8\sqrt2}3=-\tfrac{16\sqrt2}{15}$. Answer: $\dfrac{32+16\sqrt2}{15}=\dfrac{16}{15}\left(2+\sqrt2\right)\approx3.642$.
Warning The most common error is to substitute and keep the old limits: in (a) that would give $\tfrac12\big[\log u\big]_0^1$, which is not even defined.
Properties of Definite Integrals
The following properties, labelled Pโ to Pโ, turn many integrals that look impossible into easy ones. In each, $f$ is continuous on the intervals involved.
๐ Theorem 7.14 โ Properties PโโPโ
Pโ โ the variable is a dummy$\displaystyle\int_a^bf(x)\,dx=\int_a^bf(t)\,dt$
Pโ โ reversing the limits$\displaystyle\int_a^bf(x)\,dx=-\int_b^af(x)\,dx$; in particular $\displaystyle\int_a^af(x)\,dx=0$
Pโ โ splitting the interval$\displaystyle\int_a^bf(x)\,dx=\int_a^cf(x)\,dx+\int_c^bf(x)\,dx$, for any $c$ where $f$ is continuous on the intervals used
Pโ โ reflecting the interval$\displaystyle\int_a^bf(x)\,dx=\int_a^bf(a+b-x)\,dx$
Why. Pโ: both sides are $F(b)-F(a)$. Pโ: $F(b)-F(a)=-\big(F(a)-F(b)\big)$. Pโ: $\big(F(c)-F(a)\big)+\big(F(b)-F(c)\big)$ $=F(b)-F(a)$. Pโ: put $t=a+b-x$, so $dt=-dx$ and the limits $a,b$ become $b,a$; then $\displaystyle\int_a^bf(a+b-x)\,dx=-\int_b^af(t)\,dt$, which is $\displaystyle\int_a^bf(t)\,dt$ by Pโ, and Pโ finishes it. Geometrically, Pโ says that reflecting the graph in the midpoint of $[a,b]$ does not change the area.
๐ Theorem 7.15 โ Properties PโโPโ
Pโ โ from 0 to a$\displaystyle\int_0^af(x)\,dx=\int_0^af(a-x)\,dx$
Pโ โ from 0 to 2a$\displaystyle\int_0^{2a}f(x)\,dx$ $\displaystyle=\int_0^af(x)\,dx+\int_0^af(2a-x)\,dx$; so it is $\displaystyle2\int_0^af(x)\,dx$ if $f(2a-x)=f(x)$, and $0$ if $f(2a-x)=-f(x)$
Pโ โ symmetric limits$\displaystyle\int_{-a}^{a}f(x)\,dx=2\int_0^af(x)\,dx$ if $f$ is even ($f(-x)=f(x)$), and $0$ if $f$ is odd ($f(-x)=-f(x)$)
Why. Pโ is Pโ with $a=0$, $b=a$. For Pโ , split at $a$ (Pโ) and in $\int_a^{2a}f(x)\,dx$ put $x=2a-t$. For Pโ, split at $0$; in $\int_{-a}^0f(x)\,dx$ put $x=-t$ to get $\int_0^af(-t)\,dt$, which is $+\int_0^af$ for an even $f$ and $-\int_0^af$ for an odd $f$. In pictures: an even function's graph is symmetric in the $y$-axis, so the two halves have equal areas; an odd function's two halves cancel.
(a) Pโ Call the integral $I$. Replace $x$ by $\tfrac\pi2-x$: $\sin\left(\tfrac\pi2-x\right)=\cos x$ and $\cos\left(\tfrac\pi2-x\right)=\sin x$, so
$$I=\int_0^{\pi/2}\frac{\cos x}{\cos x+\sin x}\,dx$$
(a) add Adding the two forms of $I$: $2I=\displaystyle\int_0^{\pi/2}\frac{\sin x+\cos x}{\sin x+\cos x}\,dx=\int_0^{\pi/2}1\,dx=\frac\pi2$, so $I=\dfrac\pi4$ โ without ever finding an antiderivative.
(b) Pโ $x^3$ and $x\cos x$ are odd (replace $x$ by $-x$: each changes sign), so their integrals over $[-1,1]$ vanish. The constant $2$ is even: $\displaystyle\int_{-1}^12\,dx=4$. Answer: $4$.
Check (a) numerically: $I\approx0.7854=\tfrac\pi4$ โ. Graphically, the integrands $\frac{\sin x}{\sin x+\cos x}$ and $\frac{\cos x}{\sin x+\cos x}$ are mirror images in $x=\tfrac\pi4$ and add up to $1$.
โ๏ธ Example 6 โ Splitting at sign changes, and a classic Pโ problem
(a) signs $x^3-x=x(x-1)(x+1)$ is $\ge0$ on $[-1,0]$, $\le0$ on $[0,1]$ and $\ge0$ on $[1,2]$. With $F(x)=\tfrac{x^4}4-\tfrac{x^2}2$, Pโ gives
$$\big[F\big]_{-1}^{0}-\big[F\big]_0^1+\big[F\big]_1^2=\tfrac14+\tfrac14+\tfrac94=\tfrac{11}4$$
(b) Pโ Replace $x$ by $\pi-x$; $\sin(\pi-x)=\sin x$ and $\cos^2(\pi-x)=\cos^2x$:
$$I=\int_0^\pi\frac{(\pi-x)\sin x}{1+\cos^2x}\,dx$$
(b) add $2I=\pi\displaystyle\int_0^\pi\frac{\sin x}{1+\cos^2x}\,dx$. With $u=\cos x$ (limits $1$ to $-1$), this is $\pi\displaystyle\int_{-1}^{1}\frac{du}{1+u^2}=\pi\cdot\frac\pi2$. So $I=\dfrac{\pi^2}{4}$.
Why it works Pโ turned the awkward factor $x$ into the constant $\pi$, which can be taken outside. Look for Pโ whenever $x$ multiplies a function that is unchanged by $x\to a-x$.
โ ๏ธ Common mistakes when studying alone
1. Old limits after a substitution. Either change the limits with the variable (Theorem 7.13), or go back to $x$ before substituting the limits โ never a mixture.
2. Ignoring a discontinuity. $\dfrac{1}{x^2}$ is positive, yet โ$\big[-\tfrac1x\big]_{-1}^1=-2$โ. The second fundamental theorem does not apply, because $\dfrac{1}{x^2}$ is not continuous at $0\in[-1,1]$. Always check continuity on the whole interval.
3. Net area versus area. $\displaystyle\int_0^{2\pi}\sin x\,dx=0$, but the shaded area is $4$. For areas, split at the zeros (Example 6(a)).
4. Using Pโ on non-symmetric limits. โOdd function, so the integral is $0$โ is true only for $\int_{-a}^a$.
5. Keeping $+\,C$. In a definite integral the constant cancels; writing $F(b)-F(a)+C$ is wrong โ the answer is a number.
๐ Beyond the board syllabus (JEE Advanced)
In JEE Advanced 2026; not in the CBSE Class 12 syllabus or JEE Main. The rectangles of Interactive 1 give the formal definition used in JEE. Divide $[a,b]$ into $n$ equal parts of width $h=\dfrac{b-a}{n}$; then for continuous $f$
$$\int_a^bf(x)\,dx=\lim_{n\to\infty}h\big[f(a)+f(a+h)+\cdots+f(a+(n-1)h)\big]$$
Worked example. For $\displaystyle\int_0^2x^2\,dx$, $h=\tfrac2n$ and the sum is $\dfrac2n\displaystyle\sum_{r=0}^{n-1}\left(\frac{2r}{n}\right)^2=\frac{8}{n^3}\cdot\frac{(n-1)n(2n-1)}{6}$, which tends to $\tfrac{8\cdot2}{6}=\tfrac83$ โ the same as $\big[\tfrac{x^3}3\big]_0^2$. Read backwards, the definition turns limits of sums into integrals: $\displaystyle\lim_{n\to\infty}\frac1n\sum_{r=1}^{n}f\!\left(\frac rn\right)=\int_0^1f(x)\,dx$. For example $\displaystyle\lim_{n\to\infty}\sum_{r=1}^{n}\frac{n}{n^2+r^2}=\int_0^1\frac{dx}{1+x^2}=\frac\pi4$.
Exercises
I
The second fundamental theorem. Evaluate; check one or two answers with Interactive 1.
7 $u=1+\sin^2x$, $du=\sin2x\,dx$, limits $1$ to $2$: $\displaystyle\int_1^2\frac{du}u=\log2$.
8 $u=x+1$ (limits $1$ to $4$): $\displaystyle\int_1^4\left(u^{1/2}-u^{-1/2}\right)du$ $=\left[\tfrac23u^{3/2}-2u^{1/2}\right]_1^4$ $=\tfrac43+\tfrac43=\tfrac83$.
9 By Theorem 7.10, $\left[\tfrac x2\sqrt{a^2-x^2}+\tfrac{a^2}2\sin^{-1}\tfrac xa\right]_0^a=\tfrac{a^2}2\cdot\tfrac\pi2=\tfrac{\pi a^2}4$. The region is a quarter of the disc of radius $a$, whose area is $\tfrac14\pi a^2$ โ.
III
Properties PโโPโ. Use a property instead of (or before) finding an antiderivative.
5 Split at $-1$: $\displaystyle\int_{-2}^{-1}-(x+1)\,dx+\int_{-1}^{2}(x+1)\,dx$ $=\tfrac12+\tfrac92=5$.
6 Split at $1$: $\displaystyle\int_0^1(1-x)\,dx+\int_1^4(x-1)\,dx=\tfrac12+\tfrac92=5$.
7 $I=\displaystyle\int_0^\pi(\pi-x)\sin x\,dx$, so $2I=\pi\displaystyle\int_0^\pi\sin x\,dx=2\pi$ and $I=\pi$. (By parts: $\big[-x\cos x+\sin x\big]_0^\pi=\pi$ โ.)
8 Pโ: $\displaystyle\int_0^1(1-x)x^9\,dx=\tfrac1{10}-\tfrac1{11}=\tfrac1{110}$ โ much easier than expanding $(1-x)^9$.
9 $\cos^5(2\pi-x)=\cos^5x$, so Pโ gives $2\displaystyle\int_0^\pi\cos^5x\,dx$; and $\cos^5(\pi-x)=-\cos^5x$, so Pโ again (with $2a=\pi$) gives $0$. Answer: $0$.
10 Pโ with $a=\tfrac\pi4$: $\tan\left(\tfrac\pi4-x\right)=\dfrac{1-\tan x}{1+\tan x}$, so $1+\tan\left(\tfrac\pi4-x\right)=\dfrac{2}{1+\tan x}$. Then $I=\displaystyle\int_0^{\pi/4}\big(\log2-\log(1+\tan x)\big)dx$ $=\dfrac\pi4\log2-I$, so $I=\dfrac\pi8\log2$. โ
IV
Modelling with definite integrals. Units in every answer.
1Train. During a short shunting move, a train's speed is $v(t)=6t-t^2$ m/s for $0\le t\le6$ s. How far does it travel?
2Water tank. Water flows into a tank at $20+6\sqrt t$ litres per minute. How much flows in during the first $16$ minutes?
3Average temperature. The average value of $f$ on $[a,b]$ is $\dfrac{1}{b-a}\displaystyle\int_a^bf(x)\,dx$. On a summer day in Delhi the temperature is modelled by $T(t)=25+5\sin\dfrac{\pi t}{12}$ ยฐC, $t$ in hours after 6 a.m. Find the average temperature over $24$ hours and over the $12$ daytime hours (to 2 decimal places).
๐ก Show solutionsโถ
1 $v\ge0$ on $[0,6]$, so the distance is $\displaystyle\int_0^6(6t-t^2)\,dt=\left[3t^2-\tfrac{t^3}3\right]_0^6=108-72=36$ m.
2 $\displaystyle\int_0^{16}\left(20+6t^{1/2}\right)dt=\left[20t+4t^{3/2}\right]_0^{16}$ $=320+256=576$ L.
3 $\displaystyle\int_0^{24}\sin\frac{\pi t}{12}\,dt=\left[-\frac{12}\pi\cos\frac{\pi t}{12}\right]_0^{24}=0$, so the 24-hour average is $25$ ยฐC. Over $[0,12]$: $\displaystyle\int_0^{12}\sin\frac{\pi t}{12}\,dt=\frac{24}{\pi}$, so the average is $25+\dfrac{5}{12}\cdot\dfrac{24}{\pi}=25+\dfrac{10}{\pi}\approx28.18$ ยฐC.
How this is examined: the Board sets a 1-mark question on an odd function over symmetric limits, 2- and 3-mark evaluations with the fundamental theorem, and a 5-mark problem that needs Pโ (often with $x$ multiplying a trigonometric function) or splitting a modulus. JEE adds limits of sums and derivatives of integrals with variable limits.
E1.Board ยท 1 mark The value of $\displaystyle\int_{-\pi}^{\pi}x^2\sin x\,dx$ is (A) $0$ (B) $2\pi^2$ (C) $\pi^2-4$ (D) $2\pi$
๐ก Show solutionโถ
Idea $f(x)=x^2\sin x$ has $f(-x)=x^2(-\sin x)=-f(x)$: it is odd, and the limits are symmetric. By Pโ the integral is $0$.
Answer (A). (B) comes from treating the integrand as even and integrating wrongly; no calculation is needed at all.
E2.Board ยท 2 marks Evaluate $\displaystyle\int_0^{\pi/4}\tan^2x\,dx$.
๐ก Show solutionโถ
Identity $\tan^2x=\sec^2x-1$, with antiderivative $\tan x-x$. (1 mark)
E4.Board ยท 5 marks Evaluate $\displaystyle\int_0^{\pi}\frac{x}{1+\sin x}\,dx$.
๐ก Show solutionโถ
Pโ Let $I$ be the integral. Since $\sin(\pi-x)=\sin x$, $\ I=\displaystyle\int_0^\pi\frac{\pi-x}{1+\sin x}\,dx$. Adding, $2I=\pi\displaystyle\int_0^\pi\frac{dx}{1+\sin x}$. (2 marks)
Simplify Multiply above and below by $1-\sin x$:
$$\frac{1}{1+\sin x}=\frac{1-\sin x}{\cos^2x}=\sec^2x-\sec x\tan x$$
so an antiderivative is $\tan x-\sec x=\dfrac{\sin x-1}{\cos x}$. Since $\dfrac{\sin x-1}{\cos x}=\dfrac{-\cos x}{1+\sin x}$ (cross-multiply: $\sin^2x-1=-\cos^2x$), the continuous antiderivative on $[0,\pi]$ is $G(x)=-\dfrac{\cos x}{1+\sin x}$. (2 marks)
Answer $\displaystyle\int_0^\pi\frac{dx}{1+\sin x}=G(\pi)-G(0)=1-(-1)=2$, so $2I=2\pi$ and $I=\pi$. (1 mark)
Rewrite $\dfrac{n}{n^2+r^2}=\dfrac1n\cdot\dfrac{1}{1+(r/n)^2}$, so the sum is $\dfrac1n\displaystyle\sum_{r=1}^nf\!\left(\tfrac rn\right)$ with $f(x)=\dfrac{1}{1+x^2}$ (๐ box).
Answer (A): the limit is $\displaystyle\int_0^1\frac{dx}{1+x^2}=\frac\pi4$. (C) is the answer for $\sum\frac{1}{n+r}$, which gives $\int_0^1\frac{dx}{1+x}$.
E6.JEE Advanced(Numerical answer.) For $x\gt0$ let $F(x)=\displaystyle\int_1^{x^2}\log t\,dt$. Find $\dfrac{F'(e)}{e}$.
๐ก Show solutionโถ
FTC + chain rule With $A(u)=\displaystyle\int_1^u\log t\,dt$, $A'(u)=\log u$ (Theorem 7.11) and $F(x)=A(x^2)$. So $F'(x)=\log\left(x^2\right)\cdot2x=4x\log x$.
Answer $F'(e)=4e$, so $\dfrac{F'(e)}{e}=4$.
Check Directly, $F(x)=\big[t\log t-t\big]_1^{x^2}=2x^2\log x-x^2+1$, and $F'(x)=4x\log x+2x-2x=4x\log x$ โ.
๐ฌ ResearchโTeaching NexusMini project ยท about 60 minutes
Research question. How accurately can you find the area of a leaf from a few width measurements โ and is a formula fitted to the leaf's shape better than adding up strips?
Why it matters. Leaf area controls how much light a plant captures; agricultural scientists use it to estimate crop growth and water needs, and ecologists compare forests by their total leaf area. The same strip method measures the area of irregular land plots, river cross-sections (for flood flow) and wound sizes in medicine.
What to do.
Pick a large, roughly symmetric leaf (mango, banyan, teak). Lay it on graph paper, draw its midrib as the $x$-axis and measure its width $w$ every $1$ cm along the midrib. (No leaf? Use the sample widths 0, 2.1, 3.4, 4.2, 4.6, 4.7, 4.4, 3.8, 2.9, 1.7, 0 cm at $x=0,1,\dots,10$ cm.)
Estimate the area by the trapezium rule: area $\approx\sum\tfrac12(w_i+w_{i+1})\times1$.
Fit the model $w(x)=kx(L-x)$, where $L$ is the leaf's length, by matching the widest width. Compute $\displaystyle\int_0^Lkx(L-x)\,dx=\frac{kL^3}{6}$ exactly with the fundamental theorem.
Count the graph-paper squares inside the outline as a third estimate. Compare all three.
Tools: Graph paper, ruler, spreadsheet; the second fundamental theorem.
Hand in: Your table of widths, the three area estimates with percentage differences, a sketch of the model curve over your measurements, and a paragraph on where the model fits badly and why.
Going further. Measure every $0.5$ cm instead. How much does the trapezium estimate change? Why must the estimates settle down as the spacing shrinks?
Instructor note: for the sample widths the trapezium estimate is $31.8$ cmยฒ (the ends are $0$, so it is just the sum of the inner widths). Matching the maximum $4.7$ at $x=5$ with $L=10$ gives $k=0.188$ and model area $\tfrac{0.188\cdot1000}{6}\approx31.33$ cmยฒ, about $1.5\%$ below the trapezium value. Expect the square count to be within a few per cent of both.
๐งช Self-check quiz
0/5 correct
Q1. $\displaystyle\int_0^1x^2\,dx$ equals
Q2. If $f$ is continuous on $[a,b]$ and $F'=f$ there, then $\displaystyle\int_a^bf(x)\,dx$ equals
Q3. With $u=x^2$, the integral $\displaystyle\int_0^2xe^{x^2}\,dx$ becomes $\tfrac12\displaystyle\int e^u\,du$ with limits