E078 Section 2 Notes – Integration

Source: Condensed from the Section 2 Study Guide (zyBooks 2.1–2.7). Not official WGU material.

Big idea. Integration starts as a limit of Riemann sums and becomes the exact "net signed area" under a curve. The Fundamental Theorem of Calculus turns that limit into an antiderivative computation; the rest of the section is a toolkit for antiderivatives: basic formulas, substitution, exponential/log patterns, and inverse-trig patterns.

2.1 Approximating Areas

Big idea

  • Sigma notation ∑\sum means "add these up."

  • Estimate area under a curve: chop [a,b][a,b] into nn equal strips of width Δx\Delta x and add rectangle areas.

  • LnL_n: height is ff at the left edge of each strip. RnR_n: height is ff at the right edge. Any point xi∗x_i^* in each strip gives a Riemann sum; bigger nn is better.

  • Upper sum: use the max of ff on each strip. Lower sum: use the min.

Formulas

  • Δx=b−an\Delta x = \frac{b-a}{n}, xi=a+i Δxx_i = a + i\,\Delta x

  • Ln=∑i=1nf(xi−1) ΔxL_n = \sum_{i=1}^{n} f(x_{i-1})\,\Delta x

  • Rn=∑i=1nf(xi) ΔxR_n = \sum_{i=1}^{n} f(x_i)\,\Delta x

  • Riemann sum: ∑i=1nf(xi∗) Δx\sum_{i=1}^{n} f(x_i^*)\,\Delta x

  • ∑i=1nc=nc\sum_{i=1}^{n} c = nc, ∑i=1ni=n(n+1)2\sum_{i=1}^{n} i = \frac{n(n+1)}{2}, ∑i=1ni2=n(n+1)(2n+1)6\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}, ∑i=1ni3=n2(n+1)24\sum_{i=1}^{n} i^3 = \frac{n^2(n+1)^2}{4}

  • ∑(ai±bi)=∑ai±∑bi\sum (a_i \pm b_i) = \sum a_i \pm \sum b_i, ∑c ai=c∑ai\sum c\,a_i = c \sum a_i

Worked example

  • Sigma sum: ∑i=110(3i2+2)=3(385)+20=1175\sum_{i=1}^{10} (3i^2 + 2) = 3(385) + 20 = 1175, using ∑i=110i2=10⋅11⋅216=385\sum_{i=1}^{10} i^2 = \frac{10\cdot 11\cdot 21}{6} = 385.

  • Endpoints: f(x)=x3f(x) = x^3 on [0,2][0,2], n=4n = 4, Δx=0.5\Delta x = 0.5; values 0,0.125,1,3.375,80, 0.125, 1, 3.375, 8. L4=0.5(0+0.125+1+3.375)=2.25L_4 = 0.5(0 + 0.125 + 1 + 3.375) = 2.25; R4=0.5(0.125+1+3.375+8)=6.25R_4 = 0.5(0.125 + 1 + 3.375 + 8) = 6.25. Increasing ff, so L4L_4 is lower, R4R_4 is upper; true area 44 is between.

Common mistakes

  • Using all n+1n+1 points: LnL_n skips the last, RnR_n skips the first.

  • Forgetting to multiply by Δx\Delta x; writing ∑i=1nc=c\sum_{i=1}^{n} c = c instead of ncnc.

  • Assuming left means lower. True only for increasing ff; for decreasing ff, left is upper.

2.2 The Definite Integral

Big idea

  • The definite integral is the limit of Riemann sums: exact "net signed area" (above the xx-axis counts positive, below counts negative).

  • Total area uses the absolute value: ∫ab∣f(x)∣ dx\int_a^b |f(x)|\,dx.

  • Some integrals can be done with geometry (triangles, rectangles, circles).

  • Average value equals the integral divided by the interval length.

Formulas

  • ∫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

  • fave=1b−a∫abf(x) dxf_{\text{ave}} = \frac{1}{b-a}\int_a^b f(x)\,dx

  • Properties: ∫aaf=0\int_a^a f = 0; ∫baf=−∫abf\int_b^a f = -\int_a^b f; ∫ab[f±g]=∫abf±∫abg\int_a^b [f \pm g] = \int_a^b f \pm \int_a^b g; ∫abc f=c∫abf\int_a^b c\,f = c\int_a^b f; ∫abf=∫acf+∫cbf\int_a^b f = \int_a^c f + \int_c^b f

  • Comparison: if f≥gf \ge g on [a,b][a,b] then ∫abf≥∫abg\int_a^b f \ge \int_a^b g; if m≤f≤Mm \le f \le M then m(b−a)≤∫abf≤M(b−a)m(b-a) \le \int_a^b f \le M(b-a)

Worked example

  • Geometry: ∫−224−x2 dx\int_{-2}^{2} \sqrt{4 - x^2}\,dx is the top half of a circle of radius 22: area 12π(2)2=2π\frac{1}{2}\pi(2)^2 = 2\pi.

  • Net vs. total: f(x)=x−1f(x) = x - 1 on [0,4][0,4], zero at x=1x = 1. Triangle below on [0,1][0,1]: 0.50.5; above on [1,4][1,4]: 4.54.5. Net =4= 4; total =5= 5.

  • Average value: f(x)=x2f(x) = x^2 on [0,3][0,3]: ∫03x2 dx=9\int_0^3 x^2\,dx = 9, so fave=93−0=3f_{\text{ave}} = \frac{9}{3-0} = 3.

Common mistakes

  • Mixing up net and total area: for "total area," split at the zeros and use absolute values.

  • Forgetting to divide by (b−a)(b-a) for average value; flipping limits without flipping the sign.

2.3 The Fundamental Theorem of Calculus (FTC)

Big idea

  • MVT for Integrals: a continuous ff actually hits its average value at some cc in [a,b][a,b].

  • FTC Part 1: the derivative of an "area so far" function is the original function. Plug in the upper limit and multiply by its derivative (chain rule).

  • FTC Part 2: to evaluate ∫abf\int_a^b f, find any antiderivative FF and compute F(b)−F(a)F(b) - F(a).

Formulas

  • MVT: f(c)=1b−a∫abf(x) dxf(c) = \frac{1}{b-a}\int_a^b f(x)\,dx for some c∈[a,b]c \in [a,b]

  • FTC 1: ddx∫axf(t) dt=f(x)\frac{d}{dx}\int_a^x f(t)\,dt = f(x); with chain rule: ddx∫ag(x)f(t) dt=f(g(x)) g′(x)\frac{d}{dx}\int_a^{g(x)} f(t)\,dt = f(g(x))\,g'(x)

  • Two variable limits: ddx∫h(x)g(x)f(t) dt=f(g(x)) g′(x)−f(h(x)) h′(x)\frac{d}{dx}\int_{h(x)}^{g(x)} f(t)\,dt = f(g(x))\,g'(x) - f(h(x))\,h'(x)

  • FTC 2: ∫abf(x) dx=F(b)−F(a)\int_a^b f(x)\,dx = F(b) - F(a) where F′=fF' = f

Worked example

  • MVT: f(x)=x2f(x) = x^2 has average value 33 on [0,3][0,3] (from 2.2). Solve c2=3c^2 = 3: c=3≈1.73c = \sqrt{3} \approx 1.73, in [0,3][0,3].

  • FTC 1 with chain rule: ddx∫0xsin⁡t dt=sin⁡(x)⋅12x\frac{d}{dx}\int_0^{\sqrt{x}} \sin t\,dt = \sin(\sqrt{x})\cdot\frac{1}{2\sqrt{x}}.

  • Two variable limits: ddx∫x2xt3 dt=(2x)3⋅2−x3⋅1=15x3\frac{d}{dx}\int_x^{2x} t^3\,dt = (2x)^3\cdot 2 - x^3\cdot 1 = 15x^3.

  • FTC 2: ∫0πsin⁡x dx=[−cos⁡x]0π=1+1=2\int_0^{\pi} \sin x\,dx = \left[-\cos x\right]_0^{\pi} = 1 + 1 = 2.

Common mistakes

  • Forgetting the chain-rule factor g′(x)g'(x) when the limit is not just xx.

  • A variable lower limit gives a minus sign: ddx∫x5f(t) dt=−f(x)\frac{d}{dx}\int_x^5 f(t)\,dt = -f(x).

  • Writing F(a)−F(b)F(a) - F(b) in FTC 2; it is always top minus bottom. Adding +C+C to a definite integral (it cancels).

2.4 Integration Formulas and the Net Change Theorem

Big idea

  • Integrating a rate of change gives the net change: new value = old value + integral of the rate.

  • Integrating velocity gives displacement (net); integrating ∣v(t)∣|v(t)| gives total distance.

  • Even (f(−x)=f(x)f(-x) = f(x)): double the right half. Odd (f(−x)=−f(x)f(-x) = -f(x)): cancels to 00 on [−a,a][-a,a].

Formulas

  • ∫xn dx=xn+1n+1+C\int x^n\,dx = \frac{x^{n+1}}{n+1} + C (n≠−1n \ne -1)

  • ∫cos⁡x dx=sin⁡x+C\int \cos x\,dx = \sin x + C, ∫sin⁡x dx=−cos⁡x+C\int \sin x\,dx = -\cos x + C, ∫sec⁡2x dx=tan⁡x+C\int \sec^2 x\,dx = \tan x + C, ∫sec⁡xtan⁡x dx=sec⁡x+C\int \sec x \tan x\,dx = \sec x + C, ∫csc⁡2x dx=−cot⁡x+C\int \csc^2 x\,dx = -\cot x + C

  • Net change: F(b)=F(a)+∫abF′(x) dxF(b) = F(a) + \int_a^b F'(x)\,dx

  • Even: ∫−aaf=2∫0af\int_{-a}^{a} f = 2\int_0^a f. Odd: ∫−aaf=0\int_{-a}^{a} f = 0

Worked example

  • v(t)=3t−5v(t) = 3t - 5 m/s on [0,3][0,3]. Displacement: ∫03(3t−5) dt=[32t2−5t]03=13.5−15=−1.5\int_0^3 (3t - 5)\,dt = \left[\frac{3}{2}t^2 - 5t\right]_0^3 = 13.5 - 15 = -1.5 m.

  • Total distance: v=0v = 0 at t=53t = \frac{5}{3}. ∫05/3(3t−5) dt=−256\int_0^{5/3} (3t-5)\,dt = -\frac{25}{6}, ∫5/33(3t−5) dt=83\int_{5/3}^{3} (3t-5)\,dt = \frac{8}{3}, so total =256+166=416≈6.83= \frac{25}{6} + \frac{16}{6} = \frac{41}{6} \approx 6.83 m.

  • Even/odd: ∫−223x4 dx=2∫023x4 dx=1925\int_{-2}^{2} 3x^4\,dx = 2\int_0^2 3x^4\,dx = \frac{192}{5}; ∫−ππx2sin⁡x dx=0\int_{-\pi}^{\pi} x^2 \sin x\,dx = 0 since (−x)2sin⁡(−x)=−x2sin⁡x(-x)^2\sin(-x) = -x^2\sin x (odd).

Common mistakes

  • Answering "total distance" with the displacement; find where v=0v = 0 first.

  • Calling a function odd without checking f(−x)=−f(x)f(-x) = -f(x) (a constant term makes it not odd). The odd shortcut only works on a symmetric interval [−a,a][-a,a].

2.5 Substitution (u-sub)

Big idea

  • Substitution is the chain rule in reverse: find an inside function u=g(x)u = g(x) whose derivative g′(x)g'(x) is also in the integrand (up to a constant).

  • Replace everything with uu and dudu; no xx should be left.

  • For a definite integral, change the limits to uu-values and never go back to xx.

Formulas

  • ∫f(g(x)) g′(x) dx=∫f(u) du\int f(g(x))\,g'(x)\,dx = \int f(u)\,du with u=g(x)u = g(x), du=g′(x) dxdu = g'(x)\,dx

  • ∫abf(g(x)) g′(x) dx=∫g(a)g(b)f(u) du\int_a^b f(g(x))\,g'(x)\,dx = \int_{g(a)}^{g(b)} f(u)\,du

Worked example

  • Indefinite: ∫6x(3x2+4)4 dx\int 6x(3x^2+4)^4\,dx, u=3x2+4u = 3x^2 + 4, du=6x dxdu = 6x\,dx: ∫u4 du=(3x2+4)55+C\int u^4\,du = \frac{(3x^2+4)^5}{5} + C.

  • Leftover xx trick: ∫xx−1 dx\int x\sqrt{x-1}\,dx, u=x−1u = x - 1, x=u+1x = u + 1: ∫(u3/2+u1/2) du=25(x−1)5/2+23(x−1)3/2+C\int (u^{3/2} + u^{1/2})\,du = \frac{2}{5}(x-1)^{5/2} + \frac{2}{3}(x-1)^{3/2} + C.

  • Definite: ∫01x2(1+2x3)5 dx\int_0^1 x^2(1+2x^3)^5\,dx with u=1+2x3u = 1 + 2x^3, x2 dx=16dux^2\,dx = \frac{1}{6}du, limits 1→31 \to 3: 16∫13u5 du=729−136=1829\frac{1}{6}\int_1^3 u^5\,du = \frac{729 - 1}{36} = \frac{182}{9}.

Common mistakes

  • Not fixing the constant: if du=6x dxdu = 6x\,dx but you only have x dxx\,dx, use x dx=16dux\,dx = \frac{1}{6}du.

  • Keeping the old xx-limits after switching to uu.

  • Forgetting to substitute back to xx in an indefinite integral.

2.6 Integrals of Exponential and Logarithmic Functions

Big idea

  • exe^x is its own antiderivative. For other bases, divide by ln⁡a\ln a.

  • ∫1x dx=ln⁡∣x∣\int \frac{1}{x}\,dx = \ln|x| fills the gap in the power rule (n=−1n = -1).

  • Big pattern: (derivative of bottom)/(bottom) integrates to ln⁡∣bottom∣\ln|\text{bottom}|.

  • These show up in growth models (bacteria, price-demand): integrate the rate to get the change.

Formulas

  • ∫ex dx=ex+C\int e^x\,dx = e^x + C, ∫ax dx=axln⁡a+C\int a^x\,dx = \frac{a^x}{\ln a} + C

  • ∫1x dx=ln⁡∣x∣+C\int \frac{1}{x}\,dx = \ln|x| + C, ∫g′(x)g(x) dx=ln⁡∣g(x)∣+C\int \frac{g'(x)}{g(x)}\,dx = \ln|g(x)| + C

  • ∫ln⁡x dx=xln⁡x−x+C\int \ln x\,dx = x\ln x - x + C, ∫tan⁡x dx=ln⁡∣sec⁡x∣+C\int \tan x\,dx = \ln|\sec x| + C

Worked example

  • Exponentials: ∫0ln⁡2e3x dx=13e3x∣0ln⁡2=13(8−1)=73\int_0^{\ln 2} e^{3x}\,dx = \left.\frac{1}{3}e^{3x}\right|_0^{\ln 2} = \frac{1}{3}(8 - 1) = \frac{7}{3}; and ∫2x dx=2xln⁡2+C\int 2^x\,dx = \frac{2^x}{\ln 2} + C.

  • Logarithms: ∫xx2+1 dx\int \frac{x}{x^2+1}\,dx, u=x2+1u = x^2 + 1: 12∫duu=12ln⁡(x2+1)+C\frac{1}{2}\int \frac{du}{u} = \frac{1}{2}\ln(x^2+1) + C (no absolute value since x2+1>0x^2 + 1 > 0).

Common mistakes

  • ∫ax dx≠ax+1x+1\int a^x\,dx \ne \frac{a^{x+1}}{x+1}; the power rule is for xnx^n, not axa^x.

  • Dropping the absolute value in ln⁡∣x∣\ln|x|; forgetting the 1k\frac{1}{k} in ∫ekx dx=1kekx+C\int e^{kx}\,dx = \frac{1}{k}e^{kx} + C.

2.7 Integrals Resulting in Inverse Trig Functions

Big idea

  • Three shapes in the bottom: a2−u2\sqrt{a^2 - u^2} gives arcsine; a2+u2a^2 + u^2 gives arctangent; uu2−a2u\sqrt{u^2 - a^2} gives arcsecant.

  • Often a quick uu-sub comes first (u=2xu = 2x, u=x2u = x^2, u=exu = e^x).

Formulas

  • ∫dua2−u2=sin⁡−1(ua)+C\int \frac{du}{\sqrt{a^2 - u^2}} = \sin^{-1}\left(\frac{u}{a}\right) + C

  • ∫dua2+u2=1atan⁡−1(ua)+C\int \frac{du}{a^2 + u^2} = \frac{1}{a}\tan^{-1}\left(\frac{u}{a}\right) + C

  • ∫duuu2−a2=1asec⁡−1(∣u∣a)+C\int \frac{du}{u\sqrt{u^2 - a^2}} = \frac{1}{a}\sec^{-1}\left(\frac{|u|}{a}\right) + C

Worked example

  • Arctangent with uu-sub: ∫dx1+4x2\int \frac{dx}{1+4x^2}, u=2xu = 2x, du=2 dxdu = 2\,dx: 12∫du1+u2=12tan⁡−1(2x)+C\frac{1}{2}\int \frac{du}{1+u^2} = \frac{1}{2}\tan^{-1}(2x) + C.

  • Arcsine, definite: ∫03dx4−x2=sin⁡−1(x2)∣03=sin⁡−1(32)=π3\int_0^{\sqrt{3}} \frac{dx}{\sqrt{4 - x^2}} = \left.\sin^{-1}\left(\frac{x}{2}\right)\right|_0^{\sqrt{3}} = \sin^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{3}.

Common mistakes

  • Forgetting the 1a\frac{1}{a} in front of arctan and arcsec (arcsine has none).

  • Using a2a^2 instead of aa: in 19+x2\frac{1}{9+x^2}, a=3a = 3. Using degrees; inverse trig answers are in radians.

How the pieces connect

  • 2.1 builds Riemann sums; 2.2 takes n→∞n \to \infty to define the definite integral as exact net signed area.

  • 2.3 (FTC) replaces the limit with an antiderivative: ∫abf=F(b)−F(a)\int_a^b f = F(b) - F(a). The MVT for integrals comes from the average value in 2.2.

  • 2.4 supplies the basic antiderivatives FTC 2 needs and reads the integral as net change (displacement vs. total distance), with even/odd shortcuts on [−a,a][-a,a].

  • 2.5 (substitution) is the chain rule backwards; a quick uu-sub usually sets up the ln⁡\ln, exponential, or inverse-trig pattern in 2.6 and 2.7.

  • Decision flow: (1) basic formula? use it; (2) inside function whose derivative is there? uu-sub; (3) (derivative of bottom)/(bottom)? ln⁡∣bottom∣\ln|\text{bottom}|; (4) a2−u2\sqrt{a^2-u^2}, a2+u2a^2+u^2, or uu2−a2u\sqrt{u^2-a^2} in the bottom? inverse trig; (5) symmetric interval? check even/odd.

Checklist

  • 2.1: I can use the sum formulas; I can compute LnL_n and RnR_n.

  • 2.2: Net vs. total area; geometry integrals; average value.

  • 2.3: FTC 1 (with chain rule); FTC 2; MVT for integrals.

  • 2.4: Net change; total distance; even/odd.

  • 2.5: Indefinite uu-sub; definite uu-sub with new limits.

  • 2.6: exe^x and axa^x integrals; ln⁡\ln pattern.

  • 2.7: Arcsine; arctangent; arcsecant.