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

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.

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. ✓

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.

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:

  • [leftmargin=*,itemsep=1pt,topsep=1pt]
  • 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.
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):

  • [leftmargin=*,itemsep=1pt,topsep=1pt]
  • 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

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.

The Definite Integral & Its Properties

Concept
Definite integral = limit of Riemann sums = signed area
∫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

∫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

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).
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)
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

∫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).

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.

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).

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)
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

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.

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.

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)

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

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). Then:

  • [leftmargin=*,itemsep=1pt,topsep=1pt]
  • 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}.

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.