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] [ a , b ] into n n n equal strips of width Δ x \Delta x Δ x and add rectangle areas.
L n L_n L n : height is f f f at the left edge of each strip. R n R_n R n : height is f f f at the right edge. Any point x i ∗ x_i^* x i ∗ in each strip gives a Riemann sum; bigger n n n is better.
Upper sum: use the max of f f f on each strip. Lower sum: use the min.
Formulas
Δ x = b − a n \Delta x = \frac{b-a}{n} Δ x = n b − a , x i = a + i Δ x x_i = a + i\,\Delta x x i = a + i Δ x
L n = ∑ i = 1 n f ( x i − 1 ) Δ x L_n = \sum_{i=1}^{n} f(x_{i-1})\,\Delta x L n = ∑ i = 1 n f ( x i − 1 ) Δ x
R n = ∑ i = 1 n f ( x i ) Δ x R_n = \sum_{i=1}^{n} f(x_i)\,\Delta x R n = ∑ i = 1 n f ( x i ) Δ x
Riemann sum: ∑ i = 1 n f ( x i ∗ ) Δ x \sum_{i=1}^{n} f(x_i^*)\,\Delta x ∑ i = 1 n f ( x i ∗ ) Δ x
∑ i = 1 n c = n c \sum_{i=1}^{n} c = nc ∑ i = 1 n c = n c , ∑ i = 1 n i = n ( n + 1 ) 2 \sum_{i=1}^{n} i = \frac{n(n+1)}{2} ∑ i = 1 n i = 2 n ( n + 1 ) , ∑ i = 1 n i 2 = n ( n + 1 ) ( 2 n + 1 ) 6 \sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6} ∑ i = 1 n i 2 = 6 n ( n + 1 ) ( 2 n + 1 ) , ∑ i = 1 n i 3 = n 2 ( n + 1 ) 2 4 \sum_{i=1}^{n} i^3 = \frac{n^2(n+1)^2}{4} ∑ i = 1 n i 3 = 4 n 2 ( n + 1 ) 2
∑ ( a i ± b i ) = ∑ a i ± ∑ b i \sum (a_i \pm b_i) = \sum a_i \pm \sum b_i ∑ ( a i ± b i ) = ∑ a i ± ∑ b i , ∑ c a i = c ∑ a i \sum c\,a_i = c \sum a_i ∑ c a i = c ∑ a i
Worked example
Sigma sum: ∑ i = 1 10 ( 3 i 2 + 2 ) = 3 ( 385 ) + 20 = 1175 \sum_{i=1}^{10} (3i^2 + 2) = 3(385) + 20 = 1175 ∑ i = 1 10 ( 3 i 2 + 2 ) = 3 ( 385 ) + 20 = 1175 , using ∑ i = 1 10 i 2 = 10 ⋅ 11 ⋅ 21 6 = 385 \sum_{i=1}^{10} i^2 = \frac{10\cdot 11\cdot 21}{6} = 385 ∑ i = 1 10 i 2 = 6 10 ⋅ 11 ⋅ 21 = 385 .
Endpoints: f ( x ) = x 3 f(x) = x^3 f ( x ) = x 3 on [ 0 , 2 ] [0,2] [ 0 , 2 ] , n = 4 n = 4 n = 4 , Δ x = 0.5 \Delta x = 0.5 Δ x = 0.5 ; values 0 , 0.125 , 1 , 3.375 , 8 0, 0.125, 1, 3.375, 8 0 , 0.125 , 1 , 3.375 , 8 . L 4 = 0.5 ( 0 + 0.125 + 1 + 3.375 ) = 2.25 L_4 = 0.5(0 + 0.125 + 1 + 3.375) = 2.25 L 4 = 0.5 ( 0 + 0.125 + 1 + 3.375 ) = 2.25 ; R 4 = 0.5 ( 0.125 + 1 + 3.375 + 8 ) = 6.25 R_4 = 0.5(0.125 + 1 + 3.375 + 8) = 6.25 R 4 = 0.5 ( 0.125 + 1 + 3.375 + 8 ) = 6.25 . Increasing f f f , so L 4 L_4 L 4 is lower, R 4 R_4 R 4 is upper; true area 4 4 4 is between.
Common mistakes
Using all n + 1 n+1 n + 1 points: L n L_n L n skips the last, R n R_n R n skips the first.
Forgetting to multiply by Δ x \Delta x Δ x ; writing ∑ i = 1 n c = c \sum_{i=1}^{n} c = c ∑ i = 1 n c = c instead of n c nc n c .
Assuming left means lower. True only for increasing f f f ; for decreasing f f f , 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 x x x -axis counts positive, below counts negative).
Total area uses the absolute value: ∫ a b ∣ f ( x ) ∣ d x \int_a^b |f(x)|\,dx ∫ a b ∣ f ( x ) ∣ d x .
Some integrals can be done with geometry (triangles, rectangles, circles).
Average value equals the integral divided by the interval length.
Formulas
∫ a b f ( x ) d x = lim n → ∞ ∑ i = 1 n f ( x i ∗ ) Δ x \int_a^b f(x)\,dx = \lim_{n\to\infty} \sum_{i=1}^{n} f(x_i^*)\,\Delta x ∫ a b f ( x ) d x = lim n → ∞ ∑ i = 1 n f ( x i ∗ ) Δ x
f ave = 1 b − a ∫ a b f ( x ) d x f_{\text{ave}} = \frac{1}{b-a}\int_a^b f(x)\,dx f ave = b − a 1 ∫ a b f ( x ) d x
Properties: ∫ a a f = 0 \int_a^a f = 0 ∫ a a f = 0 ; ∫ b a f = − ∫ a b f \int_b^a f = -\int_a^b f ∫ b a f = − ∫ a b f ; ∫ a b [ f ± g ] = ∫ a b f ± ∫ a b g \int_a^b [f \pm g] = \int_a^b f \pm \int_a^b g ∫ a b [ f ± g ] = ∫ a b f ± ∫ a b g ; ∫ a b c f = c ∫ a b f \int_a^b c\,f = c\int_a^b f ∫ a b c f = c ∫ a b f ; ∫ a b f = ∫ a c f + ∫ c b f \int_a^b f = \int_a^c f + \int_c^b f ∫ a b f = ∫ a c f + ∫ c b f
Comparison: if f ≥ g f \ge g f ≥ g on [ a , b ] [a,b] [ a , b ] then ∫ a b f ≥ ∫ a b g \int_a^b f \ge \int_a^b g ∫ a b f ≥ ∫ a b g ; if m ≤ f ≤ M m \le f \le M m ≤ f ≤ M then m ( b − a ) ≤ ∫ a b f ≤ M ( b − a ) m(b-a) \le \int_a^b f \le M(b-a) m ( b − a ) ≤ ∫ a b f ≤ M ( b − a )
Worked example
Geometry: ∫ − 2 2 4 − x 2 d x \int_{-2}^{2} \sqrt{4 - x^2}\,dx ∫ − 2 2 4 − x 2 d x is the top half of a circle of radius 2 2 2 : area 1 2 π ( 2 ) 2 = 2 π \frac{1}{2}\pi(2)^2 = 2\pi 2 1 π ( 2 ) 2 = 2 π .
Net vs. total: f ( x ) = x − 1 f(x) = x - 1 f ( x ) = x − 1 on [ 0 , 4 ] [0,4] [ 0 , 4 ] , zero at x = 1 x = 1 x = 1 . Triangle below on [ 0 , 1 ] [0,1] [ 0 , 1 ] : 0.5 0.5 0.5 ; above on [ 1 , 4 ] [1,4] [ 1 , 4 ] : 4.5 4.5 4.5 . Net = 4 = 4 = 4 ; total = 5 = 5 = 5 .
Average value: f ( x ) = x 2 f(x) = x^2 f ( x ) = x 2 on [ 0 , 3 ] [0,3] [ 0 , 3 ] : ∫ 0 3 x 2 d x = 9 \int_0^3 x^2\,dx = 9 ∫ 0 3 x 2 d x = 9 , so f ave = 9 3 − 0 = 3 f_{\text{ave}} = \frac{9}{3-0} = 3 f ave = 3 − 0 9 = 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) ( 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 f f f actually hits its average value at some c c c in [ a , b ] [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 ∫ a b f \int_a^b f ∫ a b f , find any antiderivative F F F and compute F ( b ) − F ( a ) F(b) - F(a) F ( b ) − F ( a ) .
Formulas
MVT: f ( c ) = 1 b − a ∫ a b f ( x ) d x f(c) = \frac{1}{b-a}\int_a^b f(x)\,dx f ( c ) = b − a 1 ∫ a b f ( x ) d x for some c ∈ [ a , b ] c \in [a,b] c ∈ [ a , b ]
FTC 1: d d x ∫ a x f ( t ) d t = f ( x ) \frac{d}{dx}\int_a^x f(t)\,dt = f(x) d x d ∫ a x f ( t ) d t = f ( x ) ; with chain rule: d d x ∫ a g ( x ) f ( t ) d t = f ( g ( x ) ) g ′ ( x ) \frac{d}{dx}\int_a^{g(x)} f(t)\,dt = f(g(x))\,g'(x) d x d ∫ a g ( x ) f ( t ) d t = f ( g ( x )) g ′ ( x )
Two variable limits: d d x ∫ h ( x ) g ( x ) f ( t ) d t = 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) d x d ∫ h ( x ) g ( x ) f ( t ) d t = f ( g ( x )) g ′ ( x ) − f ( h ( x )) h ′ ( x )
FTC 2: ∫ a b f ( x ) d x = F ( b ) − F ( a ) \int_a^b f(x)\,dx = F(b) - F(a) ∫ a b f ( x ) d x = F ( b ) − F ( a ) where F ′ = f F' = f F ′ = f
Worked example
MVT: f ( x ) = x 2 f(x) = x^2 f ( x ) = x 2 has average value 3 3 3 on [ 0 , 3 ] [0,3] [ 0 , 3 ] (from 2.2). Solve c 2 = 3 c^2 = 3 c 2 = 3 : c = 3 ≈ 1.73 c = \sqrt{3} \approx 1.73 c = 3 ≈ 1.73 , in [ 0 , 3 ] [0,3] [ 0 , 3 ] .
FTC 1 with chain rule: d d x ∫ 0 x sin t d t = sin ( x ) ⋅ 1 2 x \frac{d}{dx}\int_0^{\sqrt{x}} \sin t\,dt = \sin(\sqrt{x})\cdot\frac{1}{2\sqrt{x}} d x d ∫ 0 x sin t d t = sin ( x ) ⋅ 2 x 1 .
Two variable limits: d d x ∫ x 2 x t 3 d t = ( 2 x ) 3 ⋅ 2 − x 3 ⋅ 1 = 15 x 3 \frac{d}{dx}\int_x^{2x} t^3\,dt = (2x)^3\cdot 2 - x^3\cdot 1 = 15x^3 d x d ∫ x 2 x t 3 d t = ( 2 x ) 3 ⋅ 2 − x 3 ⋅ 1 = 15 x 3 .
FTC 2: ∫ 0 π sin x d x = [ − cos x ] 0 π = 1 + 1 = 2 \int_0^{\pi} \sin x\,dx = \left[-\cos x\right]_0^{\pi} = 1 + 1 = 2 ∫ 0 π sin x d x = [ − cos x ] 0 π = 1 + 1 = 2 .
Common mistakes
Forgetting the chain-rule factor g ′ ( x ) g'(x) g ′ ( x ) when the limit is not just x x x .
A variable lower limit gives a minus sign: d d x ∫ x 5 f ( t ) d t = − f ( x ) \frac{d}{dx}\int_x^5 f(t)\,dt = -f(x) d x d ∫ x 5 f ( t ) d t = − f ( x ) .
Writing F ( a ) − F ( b ) F(a) - F(b) F ( a ) − F ( b ) in FTC 2; it is always top minus bottom. Adding + C +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)| ∣ v ( t ) ∣ gives total distance.
Even (f ( − x ) = f ( x ) f(-x) = f(x) f ( − x ) = f ( x ) ): double the right half. Odd (f ( − x ) = − f ( x ) f(-x) = -f(x) f ( − x ) = − f ( x ) ): cancels to 0 0 0 on [ − a , a ] [-a,a] [ − a , a ] .
Formulas
∫ x n d x = x n + 1 n + 1 + C \int x^n\,dx = \frac{x^{n+1}}{n+1} + C ∫ x n d x = n + 1 x n + 1 + C (n ≠ − 1 n \ne -1 n = − 1 )
∫ cos x d x = sin x + C \int \cos x\,dx = \sin x + C ∫ cos x d x = sin x + C , ∫ sin x d x = − cos x + C \int \sin x\,dx = -\cos x + C ∫ sin x d x = − cos x + C , ∫ sec 2 x d x = tan x + C \int \sec^2 x\,dx = \tan x + C ∫ sec 2 x d x = tan x + C , ∫ sec x tan x d x = sec x + C \int \sec x \tan x\,dx = \sec x + C ∫ sec x tan x d x = sec x + C , ∫ csc 2 x d x = − cot x + C \int \csc^2 x\,dx = -\cot x + C ∫ csc 2 x d x = − cot x + C
Net change: F ( b ) = F ( a ) + ∫ a b F ′ ( x ) d x F(b) = F(a) + \int_a^b F'(x)\,dx F ( b ) = F ( a ) + ∫ a b F ′ ( x ) d x
Even: ∫ − a a f = 2 ∫ 0 a f \int_{-a}^{a} f = 2\int_0^a f ∫ − a a f = 2 ∫ 0 a f . Odd: ∫ − a a f = 0 \int_{-a}^{a} f = 0 ∫ − a a f = 0
Worked example
v ( t ) = 3 t − 5 v(t) = 3t - 5 v ( t ) = 3 t − 5 m/s on [ 0 , 3 ] [0,3] [ 0 , 3 ] . Displacement: ∫ 0 3 ( 3 t − 5 ) d t = [ 3 2 t 2 − 5 t ] 0 3 = 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 ∫ 0 3 ( 3 t − 5 ) d t = [ 2 3 t 2 − 5 t ] 0 3 = 13.5 − 15 = − 1.5 m.
Total distance: v = 0 v = 0 v = 0 at t = 5 3 t = \frac{5}{3} t = 3 5 . ∫ 0 5 / 3 ( 3 t − 5 ) d t = − 25 6 \int_0^{5/3} (3t-5)\,dt = -\frac{25}{6} ∫ 0 5/3 ( 3 t − 5 ) d t = − 6 25 , ∫ 5 / 3 3 ( 3 t − 5 ) d t = 8 3 \int_{5/3}^{3} (3t-5)\,dt = \frac{8}{3} ∫ 5/3 3 ( 3 t − 5 ) d t = 3 8 , so total = 25 6 + 16 6 = 41 6 ≈ 6.83 = \frac{25}{6} + \frac{16}{6} = \frac{41}{6} \approx 6.83 = 6 25 + 6 16 = 6 41 ≈ 6.83 m.
Even/odd: ∫ − 2 2 3 x 4 d x = 2 ∫ 0 2 3 x 4 d x = 192 5 \int_{-2}^{2} 3x^4\,dx = 2\int_0^2 3x^4\,dx = \frac{192}{5} ∫ − 2 2 3 x 4 d x = 2 ∫ 0 2 3 x 4 d x = 5 192 ; ∫ − π π x 2 sin x d x = 0 \int_{-\pi}^{\pi} x^2 \sin x\,dx = 0 ∫ − π π x 2 sin x d x = 0 since ( − x ) 2 sin ( − x ) = − x 2 sin x (-x)^2\sin(-x) = -x^2\sin x ( − x ) 2 sin ( − x ) = − x 2 sin x (odd).
Common mistakes
Answering "total distance" with the displacement; find where v = 0 v = 0 v = 0 first.
Calling a function odd without checking f ( − x ) = − f ( x ) 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] [ − 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) u = g ( x ) whose derivative g ′ ( x ) g'(x) g ′ ( x ) is also in the integrand (up to a constant).
Replace everything with u u u and d u du d u ; no x x x should be left.
For a definite integral, change the limits to u u u -values and never go back to x x x .
Formulas
∫ f ( g ( x ) ) g ′ ( x ) d x = ∫ f ( u ) d u \int f(g(x))\,g'(x)\,dx = \int f(u)\,du ∫ f ( g ( x )) g ′ ( x ) d x = ∫ f ( u ) d u with u = g ( x ) u = g(x) u = g ( x ) , d u = g ′ ( x ) d x du = g'(x)\,dx d u = g ′ ( x ) d x
∫ a b f ( g ( x ) ) g ′ ( x ) d x = ∫ g ( a ) g ( b ) f ( u ) d u \int_a^b f(g(x))\,g'(x)\,dx = \int_{g(a)}^{g(b)} f(u)\,du ∫ a b f ( g ( x )) g ′ ( x ) d x = ∫ g ( a ) g ( b ) f ( u ) d u
Worked example
Indefinite: ∫ 6 x ( 3 x 2 + 4 ) 4 d x \int 6x(3x^2+4)^4\,dx ∫ 6 x ( 3 x 2 + 4 ) 4 d x , u = 3 x 2 + 4 u = 3x^2 + 4 u = 3 x 2 + 4 , d u = 6 x d x du = 6x\,dx d u = 6 x d x : ∫ u 4 d u = ( 3 x 2 + 4 ) 5 5 + C \int u^4\,du = \frac{(3x^2+4)^5}{5} + C ∫ u 4 d u = 5 ( 3 x 2 + 4 ) 5 + C .
Leftover x x x trick: ∫ x x − 1 d x \int x\sqrt{x-1}\,dx ∫ x x − 1 d x , u = x − 1 u = x - 1 u = x − 1 , x = u + 1 x = u + 1 x = u + 1 : ∫ ( u 3 / 2 + u 1 / 2 ) d u = 2 5 ( x − 1 ) 5 / 2 + 2 3 ( 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 ∫ ( u 3/2 + u 1/2 ) d u = 5 2 ( x − 1 ) 5/2 + 3 2 ( x − 1 ) 3/2 + C .
Definite: ∫ 0 1 x 2 ( 1 + 2 x 3 ) 5 d x \int_0^1 x^2(1+2x^3)^5\,dx ∫ 0 1 x 2 ( 1 + 2 x 3 ) 5 d x with u = 1 + 2 x 3 u = 1 + 2x^3 u = 1 + 2 x 3 , x 2 d x = 1 6 d u x^2\,dx = \frac{1}{6}du x 2 d x = 6 1 d u , limits 1 → 3 1 \to 3 1 → 3 : 1 6 ∫ 1 3 u 5 d u = 729 − 1 36 = 182 9 \frac{1}{6}\int_1^3 u^5\,du = \frac{729 - 1}{36} = \frac{182}{9} 6 1 ∫ 1 3 u 5 d u = 36 729 − 1 = 9 182 .
Common mistakes
Not fixing the constant: if d u = 6 x d x du = 6x\,dx d u = 6 x d x but you only have x d x x\,dx x d x , use x d x = 1 6 d u x\,dx = \frac{1}{6}du x d x = 6 1 d u .
Keeping the old x x x -limits after switching to u u u .
Forgetting to substitute back to x x x in an indefinite integral.
2.6 Integrals of Exponential and Logarithmic Functions Big idea
e x e^x e x is its own antiderivative. For other bases, divide by ln a \ln a ln a .
∫ 1 x d x = ln ∣ x ∣ \int \frac{1}{x}\,dx = \ln|x| ∫ x 1 d x = ln ∣ x ∣ fills the gap in the power rule (n = − 1 n = -1 n = − 1 ).
Big pattern: (derivative of bottom)/(bottom) integrates to ln ∣ bottom ∣ \ln|\text{bottom}| ln ∣ bottom ∣ .
These show up in growth models (bacteria, price-demand): integrate the rate to get the change.
Formulas
∫ e x d x = e x + C \int e^x\,dx = e^x + C ∫ e x d x = e x + C , ∫ a x d x = a x ln a + C \int a^x\,dx = \frac{a^x}{\ln a} + C ∫ a x d x = l n a a x + C
∫ 1 x d x = ln ∣ x ∣ + C \int \frac{1}{x}\,dx = \ln|x| + C ∫ x 1 d x = ln ∣ x ∣ + C , ∫ g ′ ( x ) g ( x ) d x = ln ∣ g ( x ) ∣ + C \int \frac{g'(x)}{g(x)}\,dx = \ln|g(x)| + C ∫ g ( x ) g ′ ( x ) d x = ln ∣ g ( x ) ∣ + C
∫ ln x d x = x ln x − x + C \int \ln x\,dx = x\ln x - x + C ∫ ln x d x = x ln x − x + C , ∫ tan x d x = ln ∣ sec x ∣ + C \int \tan x\,dx = \ln|\sec x| + C ∫ tan x d x = ln ∣ sec x ∣ + C
Worked example
Exponentials: ∫ 0 ln 2 e 3 x d x = 1 3 e 3 x ∣ 0 ln 2 = 1 3 ( 8 − 1 ) = 7 3 \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} ∫ 0 l n 2 e 3 x d x = 3 1 e 3 x 0 l n 2 = 3 1 ( 8 − 1 ) = 3 7 ; and ∫ 2 x d x = 2 x ln 2 + C \int 2^x\,dx = \frac{2^x}{\ln 2} + C ∫ 2 x d x = l n 2 2 x + C .
Logarithms: ∫ x x 2 + 1 d x \int \frac{x}{x^2+1}\,dx ∫ x 2 + 1 x d x , u = x 2 + 1 u = x^2 + 1 u = x 2 + 1 : 1 2 ∫ d u u = 1 2 ln ( x 2 + 1 ) + C \frac{1}{2}\int \frac{du}{u} = \frac{1}{2}\ln(x^2+1) + C 2 1 ∫ u d u = 2 1 ln ( x 2 + 1 ) + C (no absolute value since x 2 + 1 > 0 x^2 + 1 > 0 x 2 + 1 > 0 ).
Common mistakes
∫ a x d x ≠ a x + 1 x + 1 \int a^x\,dx \ne \frac{a^{x+1}}{x+1} ∫ a x d x = x + 1 a x + 1 ; the power rule is for x n x^n x n , not a x a^x a x .
Dropping the absolute value in ln ∣ x ∣ \ln|x| ln ∣ x ∣ ; forgetting the 1 k \frac{1}{k} k 1 in ∫ e k x d x = 1 k e k x + C \int e^{kx}\,dx = \frac{1}{k}e^{kx} + C ∫ e k x d x = k 1 e k x + C .
2.7 Integrals Resulting in Inverse Trig Functions Big idea
Three shapes in the bottom: a 2 − u 2 \sqrt{a^2 - u^2} a 2 − u 2 gives arcsine; a 2 + u 2 a^2 + u^2 a 2 + u 2 gives arctangent; u u 2 − a 2 u\sqrt{u^2 - a^2} u u 2 − a 2 gives arcsecant.
Often a quick u u u -sub comes first (u = 2 x u = 2x u = 2 x , u = x 2 u = x^2 u = x 2 , u = e x u = e^x u = e x ).
Formulas
∫ d u a 2 − u 2 = sin − 1 ( u a ) + C \int \frac{du}{\sqrt{a^2 - u^2}} = \sin^{-1}\left(\frac{u}{a}\right) + C ∫ a 2 − u 2 d u = sin − 1 ( a u ) + C
∫ d u a 2 + u 2 = 1 a tan − 1 ( u a ) + C \int \frac{du}{a^2 + u^2} = \frac{1}{a}\tan^{-1}\left(\frac{u}{a}\right) + C ∫ a 2 + u 2 d u = a 1 tan − 1 ( a u ) + C
∫ d u u u 2 − a 2 = 1 a sec − 1 ( ∣ u ∣ a ) + C \int \frac{du}{u\sqrt{u^2 - a^2}} = \frac{1}{a}\sec^{-1}\left(\frac{|u|}{a}\right) + C ∫ u u 2 − a 2 d u = a 1 sec − 1 ( a ∣ u ∣ ) + C
Worked example
Arctangent with u u u -sub: ∫ d x 1 + 4 x 2 \int \frac{dx}{1+4x^2} ∫ 1 + 4 x 2 d x , u = 2 x u = 2x u = 2 x , d u = 2 d x du = 2\,dx d u = 2 d x : 1 2 ∫ d u 1 + u 2 = 1 2 tan − 1 ( 2 x ) + C \frac{1}{2}\int \frac{du}{1+u^2} = \frac{1}{2}\tan^{-1}(2x) + C 2 1 ∫ 1 + u 2 d u = 2 1 tan − 1 ( 2 x ) + C .
Arcsine, definite: ∫ 0 3 d x 4 − x 2 = sin − 1 ( x 2 ) ∣ 0 3 = sin − 1 ( 3 2 ) = π 3 \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} ∫ 0 3 4 − x 2 d x = sin − 1 ( 2 x ) 0 3 = sin − 1 ( 2 3 ) = 3 π .
Common mistakes
Forgetting the 1 a \frac{1}{a} a 1 in front of arctan and arcsec (arcsine has none).
Using a 2 a^2 a 2 instead of a a a : in 1 9 + x 2 \frac{1}{9+x^2} 9 + x 2 1 , a = 3 a = 3 a = 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 n → ∞ to define the definite integral as exact net signed area.
2.3 (FTC) replaces the limit with an antiderivative: ∫ a b f = F ( b ) − F ( a ) \int_a^b f = F(b) - F(a) ∫ 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] [ − a , a ] .
2.5 (substitution) is the chain rule backwards; a quick u u u -sub usually sets up the ln \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? u u u -sub; (3) (derivative of bottom)/(bottom)? ln ∣ bottom ∣ \ln|\text{bottom}| ln ∣ bottom ∣ ; (4) a 2 − u 2 \sqrt{a^2-u^2} a 2 − u 2 , a 2 + u 2 a^2+u^2 a 2 + u 2 , or u u 2 − a 2 u\sqrt{u^2-a^2} u 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 L n L_n L n and R n R_n R 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 u u u -sub; definite u u u -sub with new limits.
2.6: e x e^x e x and a x a^x a x integrals; ln \ln ln pattern.
2.7: Arcsine; arctangent; arcsecant.