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c Math 152, Benjamin Aurispa Review of 6.4: The Fundamental Theorem of Calculus First, you need to make sure you refresh yourself on the antiderivatives below, since they are the foundation of ALL integrals!! Table of Some Antiderivatives: Function General Antiderivative a (a constant) ax + C xn , n 6= −1 xn+1 +C n+1 1 x ex ln |x| + C ex + C ax +C ln a ax Function 1 √ 1 − x2 1 1 + x2 General Antiderivative sin x cos x − cos x + C sin x + C sec2 x tan x + C sec x tan x csc x cot x csc2 x sec x + C − csc x + C − cot x + C arcsin x + C arctan x + C Recall that an indefinite integral is used to indicate the process of finding the most general antiderivative of f (x): Z f (x) dx = F (x) + C where F (x) is an antiderivative of f (x) (i.e. F ′ (x) = f (x)). Examples: Calculate the following general indefinite integrals. Z √ 1 3 • 4 + 7x5 + x7 − 2 dx 6x • Z 1 5 + sec x tan x − √ 2 1+x x The definite integral interval [a, b]. A B Rb a dx f (x) dx represents the NET AREA bounded between f (x) and the x−axis on the C 1 c Math 152, Benjamin Aurispa To evaluate definite integrals, we use the following: The Fundamental Theorem of Calculus, Part 2: If f is continuous on [a, b], then Z b a f (x) dx = F (b) − F (a) where F is any antiderivative of f . Examples: Evaluate the following definite integrals. • Z • Z • Z • Z 1 3 2 0 1 3 − 3 x x dx (3ex − 2x ) dx π 4 (sec2 θ + 2 sin θ) dθ 0 0 (t2 + 1)2 dt −2 2 c Math 152, Benjamin Aurispa • Z 4 1 u2 − 1 √ du u u √ 1 f (x) dx where f (x) = • 1 − x2 −1/2 Z π/2 cos x − 21 ≤ x < 0 0≤x≤ π 2 Set up the integral(s) to find the area under the curve f (x) = |x2 − 4| between x = 0 and x = 4. For a differentiable function f , if f (8) = 12 and Z 8 f ′ (x) dx = 9, what is f (−1)? −1 3 c Math 152, Benjamin Aurispa 6.5: Substitution Recall the Chain Rule for derivatives: d [g(x)]n = n[g(x)]n−1 g′ (x) dx d f (g(x)) = f ′ (g(x))g ′ (x) dx d sin(g(x)) = cos(g(x))g ′ (x) dx d g(x) e = eg(x) g′ (x) dx d g′ (x) ln(g(x)) = dx g(x) The Substitution Rule for integrals is essentially the reverse of the Chain Rule: If u = g(x) is a differentiable function, then Z Z f (g(x))g ′ (x) dx = f (u) du Examples: Z 3x2 (2 + x3 )10 dx Z te1−4t dt Z √ Z Z 2 sin x dx 1 + cos x (3 sec2 x + 6) cos(tan x + 2x) dx e5z + 2 dz e5z + 10z 4 c Math 152, Benjamin Aurispa Z 1 dx x ln(x) Z arcsin 4x √ dx 1 − 16x2 Z tan x dx Z e3x dx Z sin(11x − 5) dx Z √ 3 1 dx 9−x 5 c Math 152, Benjamin Aurispa Z 5 + 7x dx x2 + 1 Z x(x + 9)7 dx Z √ x3 dx 5x2 − 1 6 c Math 152, Benjamin Aurispa For definite integrals, there are two ways to evaluate when using substitution: Z 1 Z 9 e x √ dx x Z 4 x dx 6−x 0 4 1 p x2 x3 + 9 dx √ 7