Exam 1

Flashcard 1

Front:
What is the power rule for integration?

Back:

∫un du=un+1n+1+C,n≠−1\int u^n \, du = \frac{u^{n+1}}{n+1} + C, \quad n \neq -1∫undu=n+1un+1​+C,n=−1

Flashcard 2

Front:
What is the integral of eue^ueu?

Back:

∫eu du=eu+C\int e^u \, du = e^u + C∫eudu=eu+C

Flashcard 3

Front:
What is the integral of aua^uau (where a>0,a≠1a > 0, a \neq 1a>0,a=1)?

Back:

∫au du=auln⁡a+C\int a^u \, du = \frac{a^u}{\ln a} + C∫audu=lnaau​+C

Flashcard 4

Front:
What is the integral of 1u\frac{1}{u}u1​?

Back:

∫duu=ln⁡∣u∣+C\int \frac{du}{u} = \ln |u| + C∫udu​=ln∣u∣+C

Flashcard 5

Front:
What is the integral of sinh⁡u\sinh usinhu?

Back:

∫sinh⁡u du=cosh⁡u+C\int \sinh u \, du = \cosh u + C∫sinhudu=coshu+C

Flashcard 6

Front:
What is the integral of cosh⁡u\cosh ucoshu?

Back:

∫cosh⁡u du=sinh⁡u+C\int \cosh u \, du = \sinh u + C∫coshudu=sinhu+C

Flashcard 7

Front:
What is the formula for integration by parts?

Back:

∫u dv=uv−∫v du\int u \, dv = uv - \int v \, du∫udv=uv−∫vdu

Flashcard 8

Front:
What is the integral of sec⁡2x\sec^2 xsec2x?

Back:

∫sec⁡2x dx=tan⁡x+C\int \sec^2 x \, dx = \tan x + C∫sec2xdx=tanx+C

Flashcard 9

Front:
What is the integral of csc⁡2x\csc^2 xcsc2x?

Back:

∫csc⁡2x dx=−cot⁡x+C\int \csc^2 x \, dx = -\cot x + C∫csc2xdx=−cotx+C

Flashcard 10

Front:
What is the integral of sec⁡xtan⁡x\sec x \tan xsecxtanx?

Back:

∫sec⁡xtan⁡x dx=sec⁡x+C\int \sec x \tan x \, dx = \sec x + C∫secxtanxdx=secx+C

Flashcard 11

Front:
What is the integral of csc⁡xcot⁡x\csc x \cot xcscxcotx?

Back:

∫csc⁡xcot⁡x dx=−csc⁡x+C\int \csc x \cot x \, dx = -\csc x + C∫cscxcotxdx=−cscx+C

Flashcard 12

Front:
What is the method of partial fraction decomposition used for?

Back:
Used to integrate rational functions where the degree of the numerator is less than the degree of the denominator.

Example:

∫1(x−1)(x+2)dx\int \frac{1}{(x-1)(x+2)} dx∫(x−1)(x+2)1​dx

Decomposes into:

Ax−1+Bx+2\frac{A}{x-1} + \frac{B}{x+2}x−1A​+x+2B​

Flashcard 13

Front:
What substitution should you use for a2−x2\sqrt{a^2 - x^2}a2−x2​?

Back:
Use x=asin⁡θx = a \sin \thetax=asinθ and dx=acos⁡θdθdx = a \cos \theta d\thetadx=acosθdθ.

Flashcard 14

Front:
What substitution should you use for a2+x2\sqrt{a^2 + x^2}a2+x2​?

Back:
Use x=atan⁡θx = a \tan \thetax=atanθ and dx=asec⁡2θdθdx = a \sec^2 \theta d\thetadx=asec2θdθ.

Flashcard 15

Front:
What substitution should you use for x2−a2\sqrt{x^2 - a^2}x2−a2​?

Back:
Use x=asec⁡θx = a \sec \thetax=asecθ and dx=asec⁡θtan⁡θdθdx = a \sec \theta \tan \theta d\thetadx=asecθtanθdθ.

Flashcard 16

Front:
How do you evaluate an improper integral with an infinite bound?

Back:
Convert it into a limit:

∫a∞f(x) dx=lim⁡b→∞∫abf(x) dx\int_{a}^{\infty} f(x) \, dx = \lim_{b \to \infty} \int_{a}^{b} f(x) \, dx∫a∞​f(x)dx=b→∞lim​∫ab​f(x)dx

Flashcard 17

Front:
How do you integrate sin⁡mxcos⁡nx\sin^m x \cos^n xsinmxcosnx when one exponent is odd?

Back:

• If mmm is odd:

  • Save one sin⁡x\sin xsinx and use sin⁡2x=1−cos⁡2x\sin^2 x = 1 - \cos^2 xsin2x=1−cos2x.

  • Substitute u=cos⁡xu = \cos xu=cosx, so du=−sin⁡xdxdu = -\sin x dxdu=−sinxdx.

• If nnn is odd:

  • Save one cos⁡x\cos xcosx and use cos⁡2x=1−sin⁡2x\cos^2 x = 1 - \sin^2 xcos2x=1−sin2x.

  • Substitute u=sin⁡xu = \sin xu=sinx, so du=cos⁡xdxdu = \cos x dxdu=cosxdx.

Flashcard 18

Front:
How do you integrate sin⁡mxcos⁡nx\sin^m x \cos^n xsinmxcosnx when both exponents are even?

Back:
Use power-reduction identities:

sin⁡2x=1−cos⁡2x2,cos⁡2x=1+cos⁡2x2\sin^2 x = \frac{1 - \cos 2x}{2}, \quad \cos^2 x = \frac{1 + \cos 2x}{2}sin2x=21−cos2x​,cos2x=21+cos2x​

This reduces the integral into simpler terms that can be integrated directly.

Flashcard 19

Front:
What is a common strategy for solving ∫xex dx\int x e^x \, dx∫xexdx?

Back:
Use integration by parts with:

• u=xu = xu=x, so du=dxdu = dxdu=dx.

• dv=exdxdv = e^x dxdv=exdx, so v=exv = e^xv=ex.

∫xex dx=xex−∫ex dx=xex−ex+C\int x e^x \, dx = x e^x - \int e^x \, dx = x e^x - e^x + C∫xexdx=xex−∫exdx=xex−ex+C

Flashcard 20

Front:
What is a common strategy for solving ∫xln⁡x dx\int x \ln x \, dx∫xlnxdx?

Back:
Use integration by parts with:

• u=ln⁡xu = \ln xu=lnx, so du=dxxdu = \frac{dx}{x}du=xdx​.

• dv=xdxdv = x dxdv=xdx, so v=x22v = \frac{x^2}{2}v=2x2​.

∫xln⁡x dx=x22ln⁡x−∫x22⋅dxx\int x \ln x \, dx = \frac{x^2}{2} \ln x - \int \frac{x^2}{2} \cdot \frac{dx}{x}∫xlnxdx=2x2​lnx−∫2x2​⋅xdx​=x22ln⁡x−x24+C= \frac{x^2}{2} \ln x - \frac{x^2}{4} + C=2x2​lnx−4x2​+C

Flashcard 21

Front:
What is a common strategy for solving ∫xsin⁡x dx\int x \sin x \, dx∫xsinxdx?

Back:
Use integration by parts with:

• u=xu = xu=x, so du=dxdu = dxdu=dx.

• dv=sin⁡xdxdv = \sin x dxdv=sinxdx, so v=−cos⁡xv = -\cos xv=−cosx.

∫xsin⁡x dx=−xcos⁡x+∫cos⁡x dx\int x \sin x \, dx = -x \cos x + \int \cos x \, dx∫xsinxdx=−xcosx+∫cosxdx=−xcos⁡x+sin⁡x+C= -x \cos x + \sin x + C=−xcosx+sinx+C

Flashcard 22

Front:
How do you evaluate an improper integral with an infinite discontinuity?

Back:
Convert it into a limit at the discontinuity:

∫abf(x) dx,x=c is a discontinuity\int_{a}^{b} f(x) \, dx, \quad x = c \text{ is a discontinuity}∫ab​f(x)dx,x=c is a discontinuity

Break into two limits:

lim⁡t→c−∫atf(x) dx+lim⁡t→c+∫tbf(x) dx\lim_{t \to c^-} \int_{a}^{t} f(x) \, dx + \lim_{t \to c^+} \int_{t}^{b} f(x) \, dxt→c−lim​∫at​f(x)dx+t→c+lim​∫tb​f(x)dx

If either limit diverges, the integral is improper and divergent.