5.4 Notes - Fundamental Theorem of Calculus (Section 5.4)...

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Unformatted text preview: Fundamental Theorem of Calculus (Section 5.4) Let g( x ) = ∫ x 2 t dt Fundamental Theorem of Calculus Part 1 (FTC1): x If f is continuous on [ a, b ] then F ( x ) = ∫ f (t ) dt is continuous on [a, b] and a x differentiable on (a, b ) and F ′( x ) = d f (t ) dt = f ( x ) . dx ∫ a **This gives us a way to make continuous and differentiable functions from other continuous functions. dy Example 1: Find for each of the following functions: dx x 1) y = ∫ ( 3t 3 − 4 t 2 + 5 ) dt 3 x2 2) y = ∫ sin t dt 2 3) y = ∫ −π x cos t dt Fundamental Theorem of Calculus Part 2 (FTC2) b If f is continuous on [ a, b ] and F is an antiderivative of f, then ∫ f ( x ) dx = F (b ) − F (a ) . a b b b Notation: F (b ) − F (a ) = F ( x ) a = F ( x )]a = [ F ( x )]a This theorem gives us a nicer way to evaluate definite integrals. b To calculate ∫ f ( x ) dx : a 1) Find an antiderivative F of f. 2) Calculate F (b ) − F (a ) . Question: Why don’t we need to put +C in our antiderivative F? Example 2: Evaluate the following: 5 1) ∫ 6 dx −2 1 2) ∫ x 3 dx 0 π /2 3) ∫ sinθ dθ 0 π 4) ∫ sec 2 θ dθ 0 ⎛t ⎞ 5) ∫ ⎜ + 3 t 2 ⎟ dt ⎝t ⎠ 1 64 6) ∫ 4 3 1 dy 4−y **We call an integral improper if the integrand has a discontinuity in the interval of integration. Example 3: ∫ π −π ⎧ x −π ≤ x ≤ 0 f ( x ) dx, f ( x ) = ⎨ ⎩ sin x 0 ≤ x ≤ π Example 4: Which of the following integrals are improper? A. ∫ tan ( t + 1) dt 1 −1 B. ∫ 1 0 −2 r + 6r + 8 2 dr C . ∫ ln ( 2 s − 6 ) ds 9 4 Example 5: Evaluate ∫ 1 2 0 ⎛2 c⎞ sec z + b z + ⎜ ⎟ dz (Note: b and c are constants) ⎝ 1 − z2 ⎠ ...
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This note was uploaded on 10/02/2011 for the course MATH 2214 at Virginia Tech.

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