Integration and the Fundamental Theorem

Study Sheet

Integration and the Fundamental Theorem

Antiderivatives, Riemann sums, the definite integral, and the FTC

Antiderivatives & Indefinite Integrals

Concept
Definition
xy1−3369C = 3C = 0C = −3
Three antiderivatives of 3x2−4x+53x^2-4x+5: vertical shifts of one another, with equal slopes at every xx (dots at x=1x=1).

FF is an antiderivative of ff if F′(x)=f(x)F'(x)=f(x). Because any two antiderivatives differ by a constant, the indefinite integral collects them all:

∫f(x) dx=F(x)+C,where F′(x)=f(x).\int f(x)\,dx = F(x)+C, \qquad\text{where } F'(x)=f(x).

The +C+C is the constant of integration; never omit it on an indefinite integral.

Two functions with the same derivative have the same slope everywhere, so their difference has slope 00 everywhere and is therefore constant (that step is the Mean Value Theorem). That is why a single antiderivative plus CC describes all of them, and why an initial condition is needed to pick one.

Tip

Basic rules (each verified by differentiating the right side):

∫xn dx=xn+1n+1+C (n≠−1)∫1x dx=ln⁡∣x∣+C∫ex dx=ex+C∫cos⁡x dx=sin⁡x+C∫sin⁡x dx=−cos⁡x+C∫sec⁡2x dx=tan⁡x+C∫k dx=kx+C∫sec⁡xtan⁡x dx=sec⁡x+C\begin{aligned} \int x^{n}\,dx &= \frac{x^{n+1}}{n+1}+C \ (n\neq -1) & \int \frac{1}{x}\,dx &= \ln|x|+C\\[2pt] \int e^{x}\,dx &= e^{x}+C & \int \cos x\,dx &= \sin x + C\\[2pt] \int \sin x\,dx &= -\cos x + C & \int \sec^2 x\,dx &= \tan x + C\\[2pt] \int k\,dx &= kx + C & \int \sec x\tan x\,dx &= \sec x + C \end{aligned}

Integration is linear: ∫[af(x)+bg(x)] dx=a∫f dx+b∫g dx\displaystyle\int \big[a f(x)+b g(x)\big]\,dx = a\int f\,dx + b\int g\,dx.

Example
Worked: polynomial antiderivative

Find ∫(3x2−4x+5) dx\displaystyle\int \big(3x^2-4x+5\big)\,dx.

=3⋅x33−4⋅x22+5x+C=x3−2x2+5x+C.= 3\cdot\frac{x^3}{3} - 4\cdot\frac{x^2}{2} + 5x + C = x^3 - 2x^2 + 5x + C.

Check: ddx(x3−2x2+5x)=3x2−4x+5.\frac{d}{dx}\big(x^3-2x^2+5x\big)=3x^2-4x+5. ✓

Side note
Where it goes wrong

Forgetting the lower limit: ∫abf=F(b)−F(a)\int_a^b f = F(b) - F(a), not F(b)F(b). And for ddx∫ag(x)f(t) dt\tfrac{d}{dx}\int_a^{g(x)}f(t)\,dt the chain rule adds a factor g′(x)g'(x).

Reminder — The differentiation rules:  (xn)′=nxn−1,(fg)′=f′g+fg′,(fg)′=f′g−fg′g2,(f(g(x)))′=f′(g(x))g′(x)(x^{n})'=nx^{n-1},\quad (fg)'=f'g+fg',\quad \left(\tfrac{f}{g}\right)'=\frac{f'g-fg'}{g^{2}},\quad \big(f(g(x))\big)'=f'(g(x))g'(x)
Example
Worked: exponential and reciprocal

Find ∫(2ex+1x−x) dx\displaystyle\int \Big(2e^{x}+\frac{1}{x}-\sqrt{x}\Big)\,dx. Write x=x1/2\sqrt{x}=x^{1/2}:

=2ex+ln⁡∣x∣−x3/23/2+C=2ex+ln⁡∣x∣−23x3/2+C.= 2e^{x} + \ln|x| - \frac{x^{3/2}}{3/2} + C = 2e^{x}+\ln|x|-\frac{2}{3}x^{3/2}+C.
Try it
Quick check

Find ∫(4x3−6x2+1)dx\displaystyle\int\left(4x^3-\frac{6}{x^2}+1\right)dx.

Show the answer

Answer: x4+6x+x+Cx^4+\dfrac6x+x+C. Write −6x2=−6x−2-\dfrac{6}{x^2}=-6x^{-2}, whose antiderivative is −6⋅x−1−1=6x−1-6\cdot\dfrac{x^{-1}}{-1}=6x^{-1}. Check by differentiating: 4x3−6x−2+14x^3-6x^{-2}+1.

Riemann Sums & the Trapezoidal Rule

Concept
Approximating area with rectangles

Partition [a,b][a,b] into nn subintervals of width Δx=b−an\Delta x=\dfrac{b-a}{n}. A Riemann sum adds up rectangle areas f(xi∗) Δxf(x_i^{*})\,\Delta x, where the sample point xi∗x_i^{*} is the:

  • left endpoint (LRAM),   right endpoint (RRAM),   or midpoint (MRAM).
∫abf(x) dx≈∑i=1nf(xi∗) Δx.\int_a^b f(x)\,dx \approx \sum_{i=1}^{n} f(x_i^{*})\,\Delta x.

Area is easy for rectangles, so we cover the region with thin ones, each as tall as the function at one sample point. On each strip the function barely changes, so each rectangle is nearly the true strip of area; the total error shrinks as the strips narrow. The integral is defined as the number these sums approach.

Tip

Trapezoidal Rule. Replace each rectangle top with a slanted line (a trapezoid):

∫abf(x) dx≈Δx2[f(x0)+2f(x1)+2f(x2)+⋯+2f(xn−1)+f(xn)].\int_a^b f(x)\,dx \approx \frac{\Delta x}{2}\Big[f(x_0)+2f(x_1)+2f(x_2)+\cdots+2f(x_{n-1})+f(x_n)\Big].

The interior values are doubled; the two endpoints count once.

Tip

Over- vs. under-estimate (for a positive function):

  • Increasing: LRAM underestimates, RRAM overestimates.
  • Decreasing: LRAM overestimates, RRAM underestimates.
  • Concave up: Trapezoid overestimates, Midpoint underestimates.
  • Concave down: Trapezoid underestimates, Midpoint overestimates.
Example
Worked: right Riemann sum
xy0.511.521234
Four right-endpoint rectangles of width 0.50.5: total 3.753.75, above the true area 83\tfrac83.

Estimate ∫02x2 dx\displaystyle\int_0^2 x^2\,dx with n=4n=4 (RRAM). Here Δx=2−04=0.5\Delta x=\tfrac{2-0}{4}=0.5; right endpoints 0.5,1,1.5,20.5,1,1.5,2:

0.5[(0.5)2+(1)2+(1.5)2+(2)2]=0.5[0.25+1+2.25+4]=0.5(7.5)=3.75.0.5\big[(0.5)^2+(1)^2+(1.5)^2+(2)^2\big]=0.5\big[0.25+1+2.25+4\big]=0.5(7.5)=3.75.

Since x2x^2 is increasing, RRAM overestimates the exact value 83≈2.667\tfrac{8}{3}\approx 2.667.

Try it
Quick check

Estimate ∫03(x2+1) dx\displaystyle\int_0^3(x^2+1)\,dx with a midpoint sum using n=3n=3, and compare with the exact value 1212.

Show the answer

Answer: 11.7511.75. Δx=1\Delta x=1 and the midpoints are 0.5,1.5,2.50.5,1.5,2.5, where x2+1x^2+1 is 1.25, 3.25, 7.251.25,\ 3.25,\ 7.25. The sum is 11.7511.75, only 0.250.25 below the exact 1212.

The Definite Integral & Its Properties

Concept
Definite integral = limit of Riemann sums = signed area
xy12−112−+
∫03(x−1) dx=−12+2=32\int_0^3(x-1)\,dx=-\tfrac12+2=\tfrac32: area below the axis counts negative.
∫abf(x) dx=lim⁡n→∞∑i=1nf(xi∗) Δx.\int_a^b f(x)\,dx = \lim_{n\to\infty}\sum_{i=1}^{n} f(x_i^{*})\,\Delta x .

It equals the signed area: area above the xx-axis counts positive, area below counts negative.

Tip

Properties of definite integrals:

∫aaf dx=0∫baf dx=−∫abf dx∫ab[f±g]dx=∫abf dx±∫abg dx∫abkf dx=k∫abf dx∫acf dx=∫abf dx+∫bcf dx\begin{aligned} \int_a^a f\,dx &= 0 & \int_b^a f\,dx &= -\int_a^b f\,dx\\[2pt] \int_a^b \big[f\pm g\big]dx &= \int_a^b f\,dx \pm \int_a^b g\,dx & \int_a^b k f\,dx &= k\int_a^b f\,dx\\[2pt] \int_a^c f\,dx &= \int_a^b f\,dx + \int_b^c f\,dx && \end{aligned}
Example
Worked: definite integral by geometry
xy12324area 9
A triangle of base 33 and height 66.

∫03(2x) dx\displaystyle\int_{0}^{3}(2x)\,dx is the area of a triangle with base 33 and height 2(3)=62(3)=6: 12(3)(6)=9\tfrac12(3)(6)=9. Confirm later with the FTC.

Reminder — The Fundamental Theorem of Calculus:  ∫abf(x) dx=F(b)−F(a)where F′=f\int_a^b f(x)\,dx = F(b)-F(a) \quad\text{where } F'=f
Try it
Quick check

Use geometry or symmetry: (a) ∫04(4−x) dx\displaystyle\int_0^4(4-x)\,dx; (b) ∫−22x3 dx\displaystyle\int_{-2}^{2}x^3\,dx.

Show the answer

(a) 88: a triangle with base 44 and height 44. (b) 00: x3x^3 is odd, so the area below the axis on [−2,0][-2,0] exactly cancels the area above it on [0,2][0,2].

The Fundamental Theorem of Calculus

Concept
FTC Part 1 — accumulation functions

If ff is continuous and g(x)=∫axf(t) dtg(x)=\displaystyle\int_a^x f(t)\,dt, then

g′(x)=f(x).g'(x)=f(x).

Differentiation undoes integration. With a variable upper limit u(x)u(x), use the chain rule:

ddx∫au(x)f(t) dt=f(u(x))⋅u′(x).\frac{d}{dx}\int_a^{u(x)} f(t)\,dt = f\big(u(x)\big)\cdot u'(x).

Let A(x)A(x) be the area under ff from aa to xx. Increasing xx by a small hh adds a thin strip of width hh and height about f(x)f(x), so A(x+h)−A(x)h≈f(x)\dfrac{A(x+h)-A(x)}{h}\approx f(x), with equality in the limit. The accumulated area therefore has derivative ff: it is an antiderivative. Part 2 follows by evaluating any antiderivative at both ends.

Concept
FTC Part 2 — evaluation

If FF is any antiderivative of ff on [a,b][a,b], then

∫abf(x) dx=F(b)−F(a)=[F(x)]ab.\int_a^b f(x)\,dx = F(b)-F(a) = \Big[F(x)\Big]_a^b .
Example
Worked: FTC Part 2
xy12324area 8
The shaded trapezoid under y=2xy=2x from 11 to 33 has area 32−12=83^2-1^2=8.

∫132x dx=[x2]13=32−12=9−1=8.\displaystyle\int_1^3 2x\,dx = \Big[x^2\Big]_1^3 = 3^2-1^2 = 9-1 = 8.

Example
Worked: FTC Part 1 with the chain rule

Let g(x)=∫0x2sin⁡t dtg(x)=\displaystyle\int_0^{x^2}\sin t\,dt. Then with u=x2u=x^2, u′=2xu'=2x:

g′(x)=sin⁡(x2)⋅2x=2xsin⁡(x2).g'(x)=\sin(x^2)\cdot 2x = 2x\sin(x^2).
Try it
Quick check

(a) Evaluate ∫143x dx\displaystyle\int_1^4 3\sqrt x\,dx. (b) Find ddx∫2xet2 dt\dfrac{d}{dx}\displaystyle\int_2^x e^{t^2}\,dt.

Show the answer

(a) 1414. An antiderivative of 3x1/23x^{1/2} is 2x3/22x^{3/2}; 2(8)−2(1)=142(8)-2(1)=14. (b) ex2e^{x^2}, by FTC Part 1: differentiating the accumulation function returns the integrand at the upper limit, even though et2e^{t^2} has no elementary antiderivative.

uu-Substitution

Concept
Reversing the chain rule

If u=g(x)u=g(x) then du=g′(x) dxdu=g'(x)\,dx. Substituting turns a hard integral into a basic one:

∫f(g(x)) g′(x) dx=∫f(u) du.\int f\big(g(x)\big)\,g'(x)\,dx = \int f(u)\,du.

Indefinite: substitute back to xx at the end. Definite: either substitute back, or change the limits to uu-values and skip the back-substitution.

The chain rule says ddxF(g(x))=F′(g(x)) g′(x)\frac{d}{dx}F(g(x))=F'(g(x))\,g'(x). Reading that backwards, an integrand of the form f(g(x)) g′(x)f(g(x))\,g'(x) is the derivative of F(g(x))F(g(x)). Substitution is the bookkeeping for spotting this: naming u=g(x)u=g(x) makes the factor g′(x) dxg'(x)\,dx become dudu.

Example
Worked: indefinite uu-sub

∫2x (x2+1)3 dx\displaystyle\int 2x\,(x^2+1)^3\,dx. Let u=x2+1⇒du=2x dxu=x^2+1\Rightarrow du=2x\,dx:

∫u3 du=u44+C=(x2+1)44+C.\int u^3\,du = \frac{u^4}{4}+C = \frac{(x^2+1)^4}{4}+C.
Example
Worked: definite uu-sub, changing the limits

∫02x ex2 dx\displaystyle\int_0^2 x\,e^{x^2}\,dx. Let u=x2⇒du=2x dxu=x^2\Rightarrow du=2x\,dx, so x dx=12 dux\,dx=\tfrac12\,du. Limits: x=0→u=0x=0\to u=0, x=2→u=4x=2\to u=4:

12∫04eu du=12[eu]04=12(e4−1).\frac{1}{2}\int_0^4 e^{u}\,du = \frac{1}{2}\Big[e^{u}\Big]_0^4 = \frac{1}{2}\big(e^4-1\big).
Try it
Quick check

Evaluate ∫0π/2sin⁡3xcos⁡x dx\displaystyle\int_0^{\pi/2}\sin^3x\cos x\,dx.

Show the answer

Answer: 14\dfrac14. Let u=sin⁡xu=\sin x, du=cos⁡x dxdu=\cos x\,dx; the limits become 00 and 11. Then ∫01u3 du=14\displaystyle\int_0^1u^3\,du=\tfrac14.

Average Value & the Net Change Theorem

Concept
Average value of a function

The average value of ff on [a,b][a,b] is

favg=1b−a∫abf(x) dx.f_{\text{avg}} = \frac{1}{b-a}\int_a^b f(x)\,dx .

By the Mean Value Theorem for Integrals there is a cc in [a,b][a,b] with f(c)=favgf(c)=f_{\text{avg}}.

Reminder — The Mean Value Theorem:  f′(c)=f(b)−f(a)b−afor some c∈(a,b)f'(c)=\frac{f(b)-f(a)}{b-a} \quad\text{for some } c \in (a,b)

For finitely many values the average is the total divided by the count. For a function, the total is the integral and the count becomes the length of the interval. Geometrically, the average value is the height of the rectangle on [a,b][a,b] whose area equals the area under the curve.

Concept
Net Change Theorem

The integral of a rate of change is the net change:

∫abF′(x) dx=F(b)−F(a).\int_a^b F'(x)\,dx = F(b)-F(a).

E.g. ∫abv(t) dt\int_a^b v(t)\,dt is displacement, while ∫ab∣v(t)∣ dt\int_a^b |v(t)|\,dt is total distance.

Example
Worked: average value
xy1236height 3
The dashed rectangle of height 33 has the same area, 99, as the region under x2x^2.

Average value of f(x)=x2f(x)=x^2 on [0,3][0,3]:

13−0∫03x2 dx=13[x33]03=13⋅273=3.\frac{1}{3-0}\int_0^3 x^2\,dx = \frac{1}{3}\Big[\frac{x^3}{3}\Big]_0^3 = \frac{1}{3}\cdot\frac{27}{3} = 3.
Tip

Tip. If given a velocity v(t)v(t) and a starting position s(a)s(a), the position is s(b)=s(a)+∫abv(t) dts(b)=s(a)+\displaystyle\int_a^b v(t)\,dt. Displacement can be negative; distance never is.

Try it
Quick check

Find the average value of f(x)=xf(x)=\sqrt x on [0,4][0,4].

Show the answer

Answer: 43\dfrac43. 14∫04x1/2 dx=14⋅23⋅8=43\dfrac{1}{4}\displaystyle\int_0^4x^{1/2}\,dx=\dfrac14\cdot\dfrac23\cdot8=\dfrac43.

Going Deeper: Advanced Integration

Example
FTC Part 1 with variable bounds on BOTH ends

Find g′(x)g'(x) for g(x)=∫xx3cos⁡(t2) dtg(x)=\displaystyle\int_{x}^{x^3}\cos(t^2)\,dt. Split at any constant and apply the chain rule to each end:

g′(x)=cos⁡((x3)2)⋅3x2−cos⁡(x2)⋅1=3x2cos⁡(x6)−cos⁡(x2).g'(x)=\cos\big((x^3)^2\big)\cdot 3x^2 - \cos(x^2)\cdot 1 = 3x^2\cos(x^6)-\cos(x^2).

The lower variable limit contributes with a minus sign.

Example
Riemann sum from a table

ff is increasing; selected values below. Estimate ∫06f(x) dx\displaystyle\int_0^6 f(x)\,dx using a left sum with the given subintervals.

x0256f(x)371215\begin{array}{c|cccc} x & 0 & 2 & 5 & 6\\ \hline f(x) & 3 & 7 & 12 & 15 \end{array}

Widths 2,3,12,3,1 using left values 3,7,123,7,12:

2(3)+3(7)+1(12)=6+21+12=39.2(3)+3(7)+1(12)=6+21+12=39.

Because ff is increasing, this left sum underestimates the true integral.

Example
Definite integral by geometry (signed area)
xy−2−11212
The upper half of the circle of radius 22: area 2π2\pi.

Evaluate ∫−224−x2 dx\displaystyle\int_{-2}^{2}\sqrt{4-x^2}\,dx. The graph y=4−x2y=\sqrt{4-x^2} is the upper half of a circle of radius 22:

=12π(2)2=2π.=\tfrac12\pi(2)^2 = 2\pi.
Example
Average value application

A particle has velocity v(t)=6t−t2v(t)=6t-t^2 m/s on [0,6][0,6]. Its average velocity is

16∫06(6t−t2) dt=16[3t2−t33]06=16(108−72)=6 m/s.\frac{1}{6}\int_0^6 (6t-t^2)\,dt = \frac{1}{6}\Big[3t^2-\frac{t^3}{3}\Big]_0^6 = \frac{1}{6}\Big(108-72\Big)=6\ \text{m/s}.
Example
A limit expressed as a definite integral

Recognize the Riemann sum limit and write it as an integral:

lim⁡n→∞∑i=1n(in)2⋅1n=∫01x2 dx=13.\lim_{n\to\infty}\sum_{i=1}^{n}\Big(\frac{i}{n}\Big)^2\cdot\frac{1}{n} =\int_0^1 x^2\,dx = \frac{1}{3}.

Here Δx=1n\Delta x=\tfrac1n and the sample point xi=inx_i=\tfrac{i}{n} ranges over [0,1][0,1].

Example
Accumulation function analysis
xy1234−2−112+2−2
The graph of ff: the positive triangle adds 22 to gg, the negative one takes it back.

Let g(x)=∫0xf(t) dtg(x)=\displaystyle\int_0^x f(t)\,dt where ff is the piecewise-linear graph shown (triangle up, then down): ff rises from 00 to 22 on [0,1][0,1], returns to 00 at t=2t=2, dips to −2-2 at t=3t=3 and returns to 00 at t=4t=4. So gg climbs to g(2)=2g(2)=2 and falls back to g(4)=0g(4)=0. In general:

  • gg increases where f>0f>0 and decreases where f<0f<0.
  • gg has a local max where ff changes +→−+\to-; the max of gg is the total positive area.
  • gg is concave up where ff is increasing (g′′=f′>0g''=f'>0).
Example
Net change from a rate

Water flows into a tank at r(t)=4+2tr(t)=4+2t L/min. The amount added from t=0t=0 to t=5t=5 is

∫05(4+2t) dt=[4t+t2]05=20+25=45 liters.\int_0^5 (4+2t)\,dt = \Big[4t+t^2\Big]_0^5 = 20+25 = 45\ \text{liters}.
Try it
Quick check

For f(x)=2xf(x)=2x on [0,4][0,4] with n=4n=4 (Δx=1\Delta x=1), compute the left Riemann sum using x=0,1,2,3x=0,1,2,3.

Work it on paper before revealing the answer.

Show the answer

Answer: Left sum =1(0)+1(2)+1(4)+1(6)=12=1(0)+1(2)+1(4)+1(6)=12.

Why It Works

Let A(x)=∫axf(t) dtA(x) = \int_a^xf(t)\,dt. Then A(x+h)−A(x)A(x + h) - A(x) is the area of a thin strip of width hh under ff, which is approximately f(x)⋅hf(x)\cdot h when ff is continuous. Dividing by hh and letting h→0h \to 0 gives A′(x)=f(x)A'(x) = f(x): accumulated area grows at the rate of the function's height. Since AA and any antiderivative FF have the same derivative they differ by a constant, and A(a)=0A(a) = 0 pins it: A(x)=F(x)−F(a)A(x) = F(x) - F(a).

Tip
Reading a proof

Do not read a proof the way you read a rule. Cover the last line, ask what the previous line forces, and only then uncover it. If a step feels like magic, that is the step to work by hand with a concrete number.

Formulas, Proofs & Tips

Tip
The Fundamental Theorem of Calculus
∫abf(x) dx=F(b)−F(a)where F′=f\int_a^b f(x)\,dx = F(b)-F(a) \quad\text{where } F'=f

What it means. Integration and differentiation undo each other; a definite integral is just a difference of antiderivatives.

Example. ∫022x dx=[x2]02=4\int_0^2 2x\,dx=[x^2]_0^2=4.

Why it works. Let A(x)=∫axfA(x)=\int_a^x f. Increasing xx by hh adds a sliver of area roughly f(x)hf(x)h, so A(x+h)−A(x)h→f(x)\tfrac{A(x+h)-A(x)}{h}\to f(x) — that is, A′=fA'=f. Any two antiderivatives differ by a constant, so ∫abf=A(b)−A(a)=F(b)−F(a)\int_a^b f = A(b)-A(a)=F(b)-F(a).

Tip. Evaluate the top limit first, then subtract the bottom. The +C+C cancels in a definite integral, so you can drop it.

Tip
Substitution and integration by parts
∫f(g(x))g′(x) dx=∫f(u) du,∫u dv=uv−∫v du\int f(g(x))g'(x)\,dx=\int f(u)\,du, \qquad \int u\,dv = uv-\int v\,du

What it means. Substitution reverses the chain rule; parts reverses the product rule.

Example. ∫2x ex2 dx=ex2+C\int 2x\,e^{x^2}\,dx=e^{x^2}+C (let u=x2u=x^2, du=2x dxdu=2x\,dx).

Why it works. With u=g(x)u=g(x), du=g′(x)dxdu=g'(x)dx, so the integral literally rewrites in uu. For parts, integrate the product rule (uv)′=u′v+uv′(uv)'=u'v+uv' across the interval and rearrange.

Tip. When substituting in a definite integral, convert the limits to uu as well — then you never convert back.