Chapter 4 · Antiderivatives and Integration

4.2 The Definite Integral

Signed area under a curve — and the Fundamental Theorem that computes it with one antiderivative and two endpoints.

If $f$ is defined on $[a,b]$, the definite integral of $f$ from $a$ to $b$, written $$\int_a^b f(x)\,dx,$$ gives the area bounded by $y=f(x)$, the lines $x=a$ and $x=b$, and the $x$-axis — counting area below the axis as negative (signed area).

The notation looks like the indefinite integral, but the two are different objects: a definite integral is a number; an indefinite integral is a family of functions. The symbol $\int$, due to Leibniz, is an elongated S for "sum". The numbers $a$ and $b$ are the limits of integration (lower and upper), $f(x)$ is the integrand and $x$ the variable of integration.

📐 Properties of the definite integral
  1. $\displaystyle\int_a^b f(x)\,dx=\int_a^b f(t)\,dt=\int_a^b f(u)\,du$ — the name of the variable does not matter.
  2. $\displaystyle\int_a^a f(x)\,dx=0$.
  3. $\displaystyle\int_b^a f(x)\,dx=-\int_a^b f(x)\,dx$ — reversing the limits changes the sign.
  4. $\displaystyle\int_a^b f(x)\,dx=\int_a^c f(x)\,dx+\int_c^b f(x)\,dx$ for $c$ between $a$ and $b$.
Sums, differences and constant multiples behave as for indefinite integrals.

The Fundamental Theorem of Calculus

The Fundamental Theorem, discovered by Newton and Leibniz in the late 1600s, links differentiation and integration. It has two parts.

📐 Theorem 4.4 — Fundamental Theorem of Calculus, Part 1
If $f$ is continuous over $[a,b]$ and $F(x)=\displaystyle\int_a^x f(t)\,dt$, then $F^\prime(x)=f(x)$ over $[a,b]$.

The area function grows at a rate equal to the height of the curve. In particular, every continuous function has an antiderivative.

📐 Theorem 4.5 — Fundamental Theorem of Calculus, Part 2 (Evaluation Theorem)
If $f$ is continuous over $[a,b]$ and $F$ is any antiderivative of $f$, then $$\int_a^b f(x)\,dx=F(b)-F(a)=\Big.F(x)\Big|_a^b.$$

After approximating areas with ever more rectangles, this is almost too simple: the area of a curved region comes from evaluating one antiderivative at two endpoints and subtracting.

Interactive 1 — Area, Accumulation and the FTC

Left: the shaded signed area from $a$ to $x$ (blue above the axis, red below). Right: that area plotted as a function of $x$. Its slope at each point equals $f(x)$ — Part 1. When $x$ reaches $b$, the area equals $F(b)-F(a)$ — Part 2. Turn on rectangles to see Riemann sums close in on the same number.

off
signed area under $f$ from $a$ to $x$
accumulation $A(x) = \int_a^x f(t)\,dt$
$x$—
$f(x)$ = slope of $A$—
$A(x)$ so far—
$F(b) - F(a)$—
Riemann sum—
✏️ Example 1 — Evaluating an integral with the FTC
Use the Fundamental Theorem, Part 2, to evaluate $\displaystyle\int_{-2}^{2}\left(t^2-4\right)dt$.
Step 1 An antiderivative is $\dfrac{t^3}{3}-4t$ (power rule).
Step 2 $$\left.\left(\frac{t^3}{3}-4t\right)\right|_{-2}^{2}=\left[\frac83-8\right]-\left[-\frac83+8\right]=-\frac{32}{3}.$$
Why no $+C$ Any antiderivative works; a constant $C$ would appear in both brackets and cancel. So we take $C=0$.
Meaning The answer is negative because the parabola lies below the $t$-axis on $(-2,2)$: the region has area $\tfrac{32}{3}$, counted negatively.
✏️ Example 2 — Evaluating a definite integral with the FTC, Part 2
Evaluate $\displaystyle\int_1^9\frac{x-1}{\sqrt x}\,dx$.
Step 1 Rewrite with exponents and split: $\dfrac{x-1}{x^{1/2}}=x^{1/2}-x^{-1/2}$.
Step 2 Integrate: $\left.\left(\dfrac{x^{3/2}}{3/2}-\dfrac{x^{1/2}}{1/2}\right)\right|_1^9=\left.\left(\tfrac23x^{3/2}-2x^{1/2}\right)\right|_1^9$.
Step 3 $\left[\tfrac23(27)-2(3)\right]-\left[\tfrac23-2\right]=18-6-\tfrac23+2=\dfrac{40}{3}$.

Average Value of a Function

How do we find the average of a quantity that changes continuously — the average temperature over a day, or the average height of a curve? Start with what we know: sample $f$ at $n$ equally spaced points $x_1,\dots,x_n$ in $[a,b]$ and average the values, $$\frac{f(x_1)+f(x_2)+\cdots+f(x_n)}{n}.$$ With $\Delta x=\dfrac{b-a}{n}$ we have $\dfrac1n=\dfrac{\Delta x}{b-a}$, so the sample average is $$\frac{1}{b-a}\sum_{i=1}^{n}f(x_i)\,\Delta x,$$ which is $\dfrac1{b-a}$ times a Riemann sum. As $n\to\infty$ the Riemann sum becomes the definite integral, and the average of "infinitely many" sample values becomes an integral divided by the length of the interval.

📘 Definition — Average value of a function
If $f$ is continuous on $[a,b]$, the average value of $f$ on $[a,b]$ is $$f_{\text{ave}}=\frac{1}{b-a}\int_a^b f(x)\,dx.$$
📐 Theorem — Mean Value Theorem for Integrals
If $f$ is continuous on $[a,b]$, then there is at least one number $c$ in $[a,b]$ such that $$f(c)=f_{\text{ave}}=\frac{1}{b-a}\int_a^b f(x)\,dx,\qquad\text{equivalently}\qquad\int_a^b f(x)\,dx=f(c)\,(b-a).$$

A continuous function actually attains its average value somewhere on the interval. (It follows from the Mean Value Theorem for derivatives applied to $F(x)=\int_a^x f(t)\,dt$, or from the Intermediate Value Theorem, since $f_{\text{ave}}$ lies between the minimum and maximum of $f$.)

🧭 Geometric meaning
For $f\ge0$, the rectangle with base $[a,b]$ and height $f_{\text{ave}}$ has the same area as the region under $y=f(x)$: $f_{\text{ave}}(b-a)=\int_a^b f(x)\,dx$. Picture the region as water in a tank: once it levels out, the parts of the region that stick up above the line $y=f_{\text{ave}}$ exactly fill the gaps below it. The numbers $c$ are where the curve crosses that level line.
✏️ Example 3 — Average value and the Mean Value Theorem for Integrals
Find the average value of $f(x)=x^2-4x+5$ on $[0,4]$, and find every $c$ in $[0,4]$ with $f(c)=f_{\text{ave}}$.
Step 1 Integrate: $$\int_0^4\left(x^2-4x+5\right)dx=\left.\left(\frac{x^3}{3}-2x^2+5x\right)\right|_0^4=\frac{64}{3}-32+20=\frac{28}{3}.$$
Step 2 Divide by the length $b-a=4$: $$f_{\text{ave}}=\frac14\cdot\frac{28}{3}=\frac73\approx2.333.$$
Step 3 Solve $f(c)=\tfrac73$: $c^2-4c+5=\tfrac73\;\Rightarrow\;3c^2-12c+8=0\;\Rightarrow\;c=\dfrac{12\pm\sqrt{144-96}}{6}=2\pm\dfrac{2\sqrt3}{3}.$
Check $c\approx0.845$ and $c\approx3.155$ both lie in $[0,4]$, so there are two such values. The theorem promises at least one; it does not say there is only one.
✏️ Example 4 — Average temperature over a day
On a spring day in Nizwa the temperature $t$ hours after midnight is modelled by $T(t)=15+t-\dfrac{t^2}{24}$ °C, $0\le t\le24$. Find the average temperature over the day and the times at which the temperature equals that average.
Step 1 $$\int_0^{24}\left(15+t-\frac{t^2}{24}\right)dt=\left.\left(15t+\frac{t^2}{2}-\frac{t^3}{72}\right)\right|_0^{24}=360+288-192=456.$$
Step 2 $$T_{\text{ave}}=\frac{1}{24-0}\cdot456=19\ \text{°C}.$$ (The maximum is $T(12)=21$ °C at noon; the minimum $15$ °C at midnight.)
Step 3 Solve $T(t)=19$: $t-\dfrac{t^2}{24}=4\;\Rightarrow\;t^2-24t+96=0\;\Rightarrow\;t=12\pm4\sqrt3.$
Meaning $t\approx5.07$ h (about 5:04 a.m.) and $t\approx18.93$ h (about 6:56 p.m.). The average temperature is reached once while warming up and once while cooling down.
Interactive 3 — Average Value: Levelling the Region

The region under $f$ on $[a,b]$ is shaded. A level line — the top of a rectangle over $[a,b]$ — sloshes up and down like water and settles at $f_{\text{ave}}$. Red: parts of the region above the level; blue: gaps below it. When the line settles the two colours have equal areas ("above − gaps" reads $0$), and the dots mark the points $c$ where $f(c)=f_{\text{ave}}$.

$\int_a^b f(x)\,dx$—
$b - a$—
$f_{\text{ave}}$—
$c$ with $f(c) = f_{\text{ave}}$—
current level $h$—
area above − gaps—

Exercises

I
Evaluate each definite integral using the Fundamental Theorem of Calculus, Part 2.
  • 1 $\displaystyle\int_{-1}^{1}\left(x^2-3x\right)dx+\int_1^2\left(x^2-3x\right)dx$
  • 2 $\displaystyle\int_3^{-2}(t+2)(t-3)\,dt$
  • 3 $\displaystyle\int_3^3\left(t^2-9\right)\left(4-t^2\right)dt$
  • 4 $\displaystyle\int_{-2}^{3}\left(x^2+3x-5\right)dx$
  • 5 $\displaystyle\int_0^1x^{99}\,dx$
  • 6 $\displaystyle\int_4^8\left(4t^{5/2}-3t^{3/2}\right)dt$
  • 7 $\displaystyle\int_{1/4}^{4}\left(x^2-\frac1{x^2}\right)dx$
  • 8 $\displaystyle\int_1^2\frac{2}{x^3}\,dx$
  • 9 $\displaystyle\int_1^4\frac{1}{2\sqrt x}\,dx$
  • 10 $\displaystyle\int_1^4\frac{2-\sqrt t}{t^2}\,dt$
  • 11 $\displaystyle\int_0^{\pi/4}\sec^2\theta\,d\theta$
  • 12 $\displaystyle\int_1^{16}\frac{dt}{t^{1/4}}$
  • 13 $\displaystyle\int_0^{\pi/4}\sec\theta\tan\theta\,d\theta$
  • 14 $\displaystyle\int_0^{2\pi}\cos\theta\,d\theta$
  • 15 $\displaystyle\int_{\pi/3}^{\pi/4}\csc\theta\cot\theta\,d\theta$
  • 16 $\displaystyle\int_0^{\pi/2}\sin\theta\,d\theta$
  • 17 $\displaystyle\int_{\pi/4}^{\pi/2}\csc^2\theta\,d\theta$
See the signed area
off
signed area under $f$ from $a$ to $x$
accumulation $A(x) = \int_a^x f(t)\,dt$
$x$—
$f(x)$ = slope of $A$—
$A(x)$ so far—
$F(b) - F(a)$—
Riemann sum—
1 Property 4 combines them: $\displaystyle\int_{-1}^2\left(x^2-3x\right)dx=\left.\left(\tfrac{x^3}{3}-\tfrac{3x^2}{2}\right)\right|_{-1}^{2}=\left(\tfrac83-6\right)-\left(-\tfrac13-\tfrac32\right)=-\dfrac32$.
2 $(t+2)(t-3)=t^2-t-6$; $\left.\left(\tfrac{t^3}{3}-\tfrac{t^2}{2}-6t\right)\right|_3^{-2}=\left(-\tfrac83-2+12\right)-\left(9-\tfrac92-18\right)=\dfrac{22}{3}+\dfrac{27}{2}=\dfrac{125}{6}$. (Positive, because the limits run backwards over a region below the axis.)
3 Equal limits: the integral is $0$ (Property 2).
4 $\left.\left(\tfrac{x^3}{3}+\tfrac{3x^2}{2}-5x\right)\right|_{-2}^{3}=\left(9+\tfrac{27}{2}-15\right)-\left(-\tfrac83+6+10\right)=-\dfrac{35}{6}$.
5 $\left.\dfrac{x^{100}}{100}\right|_0^1=\dfrac1{100}$.
6 $\left.\left(\tfrac87t^{7/2}-\tfrac65t^{5/2}\right)\right|_4^8=\dfrac{35584\sqrt2-3776}{35}\approx1329.93$.
7 $\left.\left(\tfrac{x^3}{3}+\tfrac1x\right)\right|_{1/4}^{4}=\left(\tfrac{64}{3}+\tfrac14\right)-\left(\tfrac{1}{192}+4\right)=\dfrac{1125}{64}\approx17.58$.
8 $\left.\left(-\dfrac{1}{x^2}\right)\right|_1^2=-\tfrac14+1=\dfrac34$.
9 $\left.\sqrt x\right|_1^4=2-1=1$.
10 $\dfrac{2-\sqrt t}{t^2}=2t^{-2}-t^{-3/2}$; $\left.\left(-\tfrac2t+\tfrac{2}{\sqrt t}\right)\right|_1^4=\left(-\tfrac12+1\right)-(-2+2)=\dfrac12$.
11 $\left.\tan\theta\right|_0^{\pi/4}=1$.
12 $\left.\tfrac43t^{3/4}\right|_1^{16}=\tfrac43(8-1)=\dfrac{28}{3}$.
13 $\left.\sec\theta\right|_0^{\pi/4}=\sqrt2-1\approx0.414$.
14 $\left.\sin\theta\right|_0^{2\pi}=0$ — the positive and negative areas cancel.
15 $\left.(-\csc\theta)\right|_{\pi/3}^{\pi/4}=-\sqrt2+\dfrac{2}{\sqrt3}=\dfrac{2\sqrt3}{3}-\sqrt2\approx-0.260$.
16 $\left.(-\cos\theta)\right|_0^{\pi/2}=0+1=1$.
17 $\left.(-\cot\theta)\right|_{\pi/4}^{\pi/2}=0+1=1$.
II
Average value and the Mean Value Theorem for Integrals.
  • 1 Find the average value of $f(x)=3x^2+2x$ on $[1,3]$ and the value of $c$ guaranteed by the Mean Value Theorem for Integrals.
  • 2 Find the average value of $f(x)=\sqrt x$ on $[0,9]$ and find $c$ with $f(c)=f_{\text{ave}}$.
  • 3 Find the average value of $f(x)=\cos x$ on $\left[0,\tfrac{\pi}{2}\right]$ and the corresponding $c$.
  • 4 Find the average value of $f(x)=\dfrac{1}{x^2}$ on $[1,4]$ and the corresponding $c$.
  • 5 Population. A town's population $t$ years after 2015 is $P(t)=50+4t+0.3t^2$ thousand. Find the average population over $0\le t\le10$ and the year in which the population equalled that average.
  • 6 Average velocity. A particle moves with velocity $v(t)=t^2-4t+3$ m/s for $0\le t\le6$. Find its average velocity and show that it equals displacement ÷ time. When is $v(t)$ equal to the average velocity?
  • 7 Conceptual. The average value of a continuous function $f$ on $[2,7]$ is $4$. Find $\displaystyle\int_2^7f(x)\,dx$ and explain why $f$ must take the value $4$ somewhere in $[2,7]$.
  • 8 Find $k>0$ so that the average value of $f(x)=x^2$ on $[0,k]$ is $12$.
Level the region
$\int_a^b f(x)\,dx$—
$b - a$—
$f_{\text{ave}}$—
$c$ with $f(c) = f_{\text{ave}}$—
current level $h$—
area above − gaps—
1 $\displaystyle\int_1^3\left(3x^2+2x\right)dx=\left.\left(x^3+x^2\right)\right|_1^3=36-2=34$, so $f_{\text{ave}}=\dfrac{34}{2}=17$. Then $3c^2+2c-17=0$ gives $c=\dfrac{-2\pm\sqrt{4+204}}{6}=\dfrac{-1\pm2\sqrt{13}}{3}$; only $c=\dfrac{-1+2\sqrt{13}}{3}\approx2.070$ lies in $[1,3]$.
2 $\displaystyle\int_0^9x^{1/2}\,dx=\left.\tfrac23x^{3/2}\right|_0^9=18$, so $f_{\text{ave}}=\dfrac{18}{9}=2$. $\sqrt c=2\Rightarrow c=4$.
3 $\displaystyle\int_0^{\pi/2}\cos x\,dx=\left.\sin x\right|_0^{\pi/2}=1$, so $f_{\text{ave}}=\dfrac{1}{\pi/2}=\dfrac{2}{\pi}\approx0.637$. $\cos c=\dfrac2\pi\Rightarrow c=\cos^{-1}\!\left(\dfrac2\pi\right)\approx0.881$.
4 $\displaystyle\int_1^4x^{-2}\,dx=\left.\left(-\tfrac1x\right)\right|_1^4=-\tfrac14+1=\tfrac34$, so $f_{\text{ave}}=\dfrac{3/4}{3}=\dfrac14$. $\dfrac1{c^2}=\dfrac14\Rightarrow c=2$ (taking $c>0$).
5 $\displaystyle\int_0^{10}\left(50+4t+0.3t^2\right)dt=\left.\left(50t+2t^2+0.1t^3\right)\right|_0^{10}=500+200+100=800$, so $P_{\text{ave}}=\dfrac{800}{10}=80$ thousand people. $50+4c+0.3c^2=80\Rightarrow3c^2+40c-300=0\Rightarrow c=\dfrac{-20+10\sqrt{13}}{3}\approx5.35$, i.e. during 2020 (about $5.35$ years after 2015).
6 Displacement $=\displaystyle\int_0^6\left(t^2-4t+3\right)dt=\left.\left(\tfrac{t^3}{3}-2t^2+3t\right)\right|_0^6=72-72+18=18$ m, so $v_{\text{ave}}=\dfrac{18}{6}=3$ m/s $=\dfrac{\text{displacement}}{\text{time}}$. $t^2-4t+3=3\Rightarrow t(t-4)=0\Rightarrow t=0$ or $t=4$ s.
7 $\displaystyle\int_2^7f(x)\,dx=f_{\text{ave}}(b-a)=4(7-2)=20$. Because $f$ is continuous, the Mean Value Theorem for Integrals gives some $c\in[2,7]$ with $f(c)=f_{\text{ave}}=4$.
8 $f_{\text{ave}}=\dfrac1k\displaystyle\int_0^kx^2\,dx=\dfrac1k\cdot\dfrac{k^3}{3}=\dfrac{k^2}{3}$. Setting $\dfrac{k^2}{3}=12$ gives $k^2=36$, so $k=6$.
III
Using the properties of the definite integral. Suppose $\displaystyle\int_1^4f(x)\,dx=6$, $\displaystyle\int_1^6f(x)\,dx=10$ and $\displaystyle\int_1^4g(x)\,dx=-2$. Find:
  • 1 $\displaystyle\int_4^6f(x)\,dx$
  • 2 $\displaystyle\int_1^4\big(3f(x)-2g(x)\big)\,dx$
  • 3 $\displaystyle\int_4^1f(x)\,dx$
  • 4 $\displaystyle\int_1^4\big(f(x)+5\big)\,dx$
  • 5 the average value of $f$ on $[1,6]$
1 Property 4: $\displaystyle\int_1^6f=\int_1^4f+\int_4^6f$, so $\displaystyle\int_4^6f(x)\,dx=10-6=4$.
2 Constant multiples and differences: $3\displaystyle\int_1^4f-2\int_1^4g=3(6)-2(-2)=18+4=22$.
3 Property 3 (reversing the limits): $\displaystyle\int_4^1f(x)\,dx=-\int_1^4f(x)\,dx=-6$.
4 $\displaystyle\int_1^4f(x)\,dx+\int_1^45\,dx=6+5(4-1)=6+15=21$.
5 $f_{\text{ave}}=\dfrac{1}{6-1}\displaystyle\int_1^6f(x)\,dx=\dfrac{10}{5}=2$.
IV
Areas, net area and total area, and Riemann sums.
  • 1 Use geometry (not antiderivatives) to evaluate $\displaystyle\int_{-3}^{3}\sqrt{9-x^2}\,dx$.
  • 2 Use geometry to evaluate $\displaystyle\int_0^4|x-2|\,dx$.
  • 3 For $f(x)=x-2$ on $[0,6]$, find the net signed area $\displaystyle\int_0^6(x-2)\,dx$ and the total area between the graph and the $x$-axis.
  • 4 For $f(x)=x^2-1$ on $[0,2]$, find the net signed area and the total area.
  • 5 Approximate $\displaystyle\int_0^2x^2\,dx$ with $n=4$ subintervals using the left, right and midpoint Riemann sums, and compare with the exact value.
  • 6 Conceptual. Without computing an antiderivative, explain why $\displaystyle\int_{-2}^{2}x^3\,dx=0$. Then find the total area between $y=x^3$ and the $x$-axis on $[-2,2]$.
Net versus total area
off
signed area under $f$ from $a$ to $x$
accumulation $A(x) = \int_a^x f(t)\,dt$
$x$—
$f(x)$ = slope of $A$—
$A(x)$ so far—
$F(b) - F(a)$—
Riemann sum—
1 $y=\sqrt{9-x^2}$ is the upper half of the circle $x^2+y^2=9$ of radius $3$. The integral is the area of the semicircle: $\tfrac12\pi(3)^2=\dfrac{9\pi}{2}\approx14.14$.
2 The graph is a "V" with vertex $(2,0)$: two triangles, each with base $2$ and height $2$. Area $=2\cdot\tfrac12(2)(2)=4$.
3 Below the axis on $[0,2]$: a triangle of area $\tfrac12(2)(2)=2$. Above on $[2,6]$: a triangle of area $\tfrac12(4)(4)=8$. Net area $=8-2=6$ (check: $\left.\left(\tfrac{x^2}{2}-2x\right)\right|_0^6=18-12=6$). Total area $=8+2=10$.
4 $x^2-1<0$ on $[0,1)$ and $>0$ on $(1,2]$. $\displaystyle\int_0^1\left(x^2-1\right)dx=\tfrac13-1=-\tfrac23$ and $\displaystyle\int_1^2\left(x^2-1\right)dx=\left(\tfrac83-2\right)-\left(\tfrac13-1\right)=\tfrac43$. Net area $=-\tfrac23+\tfrac43=\tfrac23$; total area $=\tfrac23+\tfrac43=2$.
5 $\Delta x=\tfrac12$. Left: $\tfrac12\left(0+\tfrac14+1+\tfrac94\right)=\tfrac74=1.75$. Right: $\tfrac12\left(\tfrac14+1+\tfrac94+4\right)=\tfrac{15}{4}=3.75$. Midpoint: $\tfrac12\left(\tfrac1{16}+\tfrac9{16}+\tfrac{25}{16}+\tfrac{49}{16}\right)=\tfrac{21}{8}=2.625$. Exact: $\left.\tfrac{x^3}{3}\right|_0^2=\tfrac83\approx2.667$. $f$ is increasing, so the left sum underestimates and the right sum overestimates; the midpoint sum is closest.
6 $x^3$ is odd: the region on $[-2,0]$ is the mirror image, below the axis, of the region on $[0,2]$, so the signed areas cancel and the integral is $0$. Total area $=2\displaystyle\int_0^2x^3\,dx=2\cdot\left.\tfrac{x^4}{4}\right|_0^2=2\cdot4=8$.
🔬 Research–Teaching NexusResearch-oriented · optional, about 30 minutes

Research question. What is the average temperature in Nizwa over one day, and does the method matter?

Why it matters. Meteorologists and energy engineers compute daily averages to estimate cooling needs and solar-panel output; the averaging method affects the result.

What to do.

  1. Record the temperature every 3 hours from a weather app, or use sample data (°C) at 0, 3, 6, …, 24 h: 27, 25, 24, 30, 36, 38, 34, 30, 28.
  2. Compute the average value $\frac{1}{24}\int_0^{24}T(t)\,dt$ using the trapezoid rule.
  3. Compare with the simple mean of the readings and with (max + min)/2.
  4. Optionally fit a model in Desmos and integrate it.

Tools: Weather app, spreadsheet or Desmos

Hand in: A half-page note: the three averages, the method you trust most, and why.

Going further. Anywhere: repeat for your own city with hourly readings.

Instructor note: Students should notice that the trapezoid rule gives half weight to endpoints, while the simple mean counts both midnights fully.

🧪 Self-check quiz

0/5 correct

Q1. A definite integral $\displaystyle\int_a^bf(x)\,dx$ is:

Q2. $\displaystyle\int_5^2f(x)\,dx=$

Q3. If $F(x)=\displaystyle\int_0^x\cos t\,dt$, then $F^\prime(x)=$

Q4. $\displaystyle\int_{-2}^{2}\left(t^2-4\right)dt=$

Q5. The average value of $f(x)=x^2$ on $[0,3]$ is:

← 4.1 Antiderivatives4.3 Integration by Substitution →