MATH 132 MIDTERM 1 REVIEW. 1. Compute derivatives of integrals Theorem 1 (The Fundamental Theorem of Calculus). Suppose f is continuous on

MATH 132 MIDTERM 1 REVIEW 1. Compute derivatives of integrals Theorem 1 (The Fundamental Theorem of Calculus). Suppose f is continuous on [a, b]. Rx ...
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MATH 132 MIDTERM 1 REVIEW

1. Compute derivatives of integrals Theorem 1 (The Fundamental Theorem of Calculus). Suppose f is continuous on [a, b]. Rx (1) If g(x) = a f (t) dt, then g 0 (x) = f (x) Rb (2) a f (x) dx = F (b) − F (a), where F is any antiderivative of f , that is, F 0 = f . Remark 2. Suppose that u(x), v(x) are differentiable functions, then Z u(x) du(x) d f (t) dt = f (u(x)) dx a dx Z a d dv(x) f (t) dt = −f (v(x)) dx v(x) dx Z u(x) Z u(x) Z a d d du(x) dv(x) d f (t) dt = f (t) dt + f (t) dt = f (u(x)) − f (v(x)) dx v(x) dx a dx v(x) dx dx 2. Net change problems Theorem 3 (Net change theorem). The integral of a rate of change is the net change: Z b F 0 (x) dx = F (b) − F (a) a

3. Compute areas and volumes Theorem 4. The area A of the region bounded by the curves y = f (x), y = g(x), and the lines x = a, x = b, where f and g are continuous and f (x) ≥ g(x) for all x in [a, b], is Z b A= [f (x) − g(x)] dx a

Definition 5 (Definition of volume). Let S be a solid that lies between x = a and x = b. If the cross-sectional area of S in the plabe Px , through x and perpendicular to the x-axis, is A(x), where A is continuous function, then the volume of S is Z b V = A(x) dx a

Remark 6. In general, 1

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(1) the region is rotated about the x-axis, then the volume is Z b V = π[(outer function)2 − (inner function)2 ] dx a Z b π[(outer radius)2 − (inner radius)2 ] dx = a

(2) the region is rotated about the y-axis, then the volume is Z b π[(outer function)2 − (inner function)2 ] dy V = a Z b π[(outer radius)2 − (inner radius)2 ] dy = a

(3) the region is rotated about the y = r, then the volume is Z b V = π[(outer function − r)2 − (inner function − r)2 ] dx a Z b π[(outer radius)2 − (inner radius)2 ] dx = a

(4) the region is rotated about the x = r, then the volume is Z b V = π[(outer function − r)2 − (inner function − r)2 ] dy a Z b = π[(outer radius)2 − (inner radius)2 ] dy a

4. Table of indefinite integrals

Don’t forget 0 + C 0!!! Z

Z k dx = kx + C

Z Z

1 dx = ln|x| + C x ex dx = ex + C

Z sin x dx = − cos x + C Z

sec2 x dx = tan x + C

Z sec x tan x dx = sec x + C Z



1 dx = arcsin x + C 1 − x2

1 xn+1 + C (n 6= −1) xn dx = n+1 Z 1 dx = arctan x + C 1 + x2 Z 1 x ax dx = a +C ln a Z cos x dx = sin x + C Z csc2 x dx = − cot x + C Z csc x cot x dx = − csc x + C Z 1 −√ dx = arccos x + C 1 − x2

MIDTERM 1 REVIEW

Z

3

Z csc x dx = ln|csc x − cot x| + C Z cot x dx = ln|sin x| + C

sec x dx = ln|sec x + tan x| + C Z tan x dx = ln|sec x| + C Remark 7. Note that

R

f (x)g(x) dx 6=

R

f (x) dx ·

R

g(x) dx!!!

5. Integration methods 5.1. U substitution. Theorem 8 (The substitution rule for indefinite integrals). If u = g(x) is a differentiable function whose range is an interval I and f is continuous on I, then Z Z 0 f (g(x))g (x) dx = f (u) du Theorem 9 (The substitution rule for definite integrals). If g 0 is continuous on [a, b] and f is continuous on the range of u = g(x), then Z g(b) Z b 0 f (u) du f (g(x))g (x) dx = g(a)

a

5.2. Integration by parts. Theorem 10 (The formula for integration by parts for indefinite integral). Z Z u dv = uv − v du Theorem 11 (The formula for integration by parts for definite integral). Z b b Z b 0 g(x)f 0 (x) dx f (x)g (x) dx = f (x)g(x) − a

a

a

Theorem 12 (ILATE rule). Whichever function comes first in the following list should be u: I - Inverse trigonometric functions: arctan(x), arcsin(x) etc. L - Logarithmic Functions: ln(x), log(2x) etc. A - Algebraic functions: x2 , x50 etc. T - Trigonometric functions: sin(x), tan(x) etc. E - Exponential functions: ex , 2x etc. 5.3. Trigonometric integrals. R • Strategy for evaluating sinm x cosn x dx (1) ODD COSINE (a) Factor out one power of cosine. (b) Replace the remaining even powers of cosine using cos2 x = 1 − sin2 x (c) Let u = sin x and integrate. (2) ODD SINE

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(a) Factor out one power of sine. (b) Replace the remaining even powers of sine using sin2 x = 1 − cos2 x (c) Let u = cos x and integrate. (3) ODD SINE and COSINE You can use either of the above methods. (4) EVEN SINE and COSINE You will use the half angle identities. 1 1 1 sin2 x = (1 − cos 2x) cos2 x = (1 + cos 2x) sin x cos x = sin 2x 2 2 2 R m n • Strategy for evaluating tan x sec x dx (1) EVEN SECANT (a) Factor out a sec2 x. (b) Replace the remaining even powers of secant using sec2 x = 1 + tan2 x (c) Let u = tan x and integrate. (2) ODD TANGENT (a) Factor out a sec x tan x. (b) Replace the remaining powers of tangent using tan2 x = sec2 x − 1 (c) Let u = sec x and integrate. 5.4. Trigonometric substitution. Proposition 13 (Trigonometric Identities). 1 1 sin x cos x 1 sec x = cot x = tan x = cot x = sin x cos x tan x cos x sin x sin2 x + cos2 x = 1 tan2 x + 1 = sec2 x 1 + cot2 x = csc2 x 1 sin x cos x = sin 2x cos 2x = cos2 x − sin2 x 2 1 1 2 sin x = (1 − cos 2x) cos2 x = (1 + cos 2x) 2 2

csc x =

Expression Substitution Identity √ a2 − x 2 x = a sin θ 1 − sin2 θ = cos2 θ √ a2 + x 2 x = a tan θ 1 + tan2 θ = sec2 θ √ x 2 − a2 x = a sec θ sec2 θ − 1 = tan2 θ

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5.5. Integration of rational functions. Theorem 14 (Strategy for integration of rational functions by partial fractions). Consider a rational function polynomial P (x) = f (x) = Q(x) polynomial (1) Step 1: check if deg(P ) < deg(Q). If YES, go to Step 2; if NO, perform the long division. (2) Step 2: check if the denominator Q(x) is factored as far as possible. If YES, go to Step 3; if NO, factor Q(x). (3) Step 3: express the proper rational function as a sum of partial fractions of the form (see the following table for details) A Ax + B or i 2 (ax + b) (ax + bx + c)j Factor in denominator Term in partial fraction decomposition A ax+b

ax + b A1 ax+b

(ax + b)r

A2 (ax+b)2

+ ··· +

Ar (ax+b)r

Ax+B ax2 +bx+c

ax2 + bx + c (ax2 + bx + c)r

+

A1 x+B1 ax2 +bx+c

+

A2 x+B2 (ax2 +bx+c)2

+ ··· +

Ar x+Br (ax2 +bx+c)r

Remark 15. In this exam, you only need to know how to proceed step by step above and how to compute the first case. 6. Strategy for integration 6.1. Basic methods. • U substitution: integrand includes composite functions • Integration by parts: integrand is a product of functions • Trigonometric integrals: integrand is a product of trig. functions • Trigonometric substitution: integrand includes even roots of certain polynomials • Rational function: integrand is a rational function 6.2. How to find a suitable strategy to evaluate an integral. (1) Simplify the integrand: factoring, completing the square and trig. identities (2) Always try U-sub first. (3) Look for suitable technique based on the form of integrand: • Products of functions: integration by parts • Products of trig. functions: trig. integrals • Even roots of certain polynomials: trig. substitution • Rational functions: partial fractions