Workshop_Solutions_5

2 4 mathematical methods for physics 1 solutions to

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Unformatted text preview: rd step, and where c ∈ R is an arbitrary constant. 2 4 Mathematical Methods for Physics 1, Solutions to Workshop Sheet 5 4. Use integration by substitution to evaluate the integral 1 I1 = 0 1 dx. 2x+5 Solution: Making the substitution dy =2 dx y = y (x) = 2 x + 5, ⇒ dx = 1 dy, 2 with the new boundaries y (0) = 2 × 0 + 5 = 5 and y (1) = 2 × 1 + 5 = 7, gives that 1 I1 = 0 1 dx = 2x+5 7 5 1 11 dy = ln |y | y2 2 7 5 = 1 ln(7/5) ln(7) − ln(5) = . 2 2 2 5. Evaluate the definite integral 1 J= 0 √ x dx. 1+x Solution: To evaluate J , we integrate by parts with f (x) = 2 (1 + x)1/2 f ′ (x) = (1 + x)−1/2 , g (x) = x, and g ′ (x) = 1. Then we obtain 1 √ J= 0 1 = 0 x dx = 1+x 1 x (1 + x)−1/2 dx 0 f ′ (x) g (x) dx = f (x) g (x)|1 − 0 1 f (x) g ′(x) dx 0 1 √ √ 2 (1 + x)3/2 4 = 2 2 − (1 + x)3/2 2 (1 + x)1/2 dx = 2 2 − 3/2 3 0 0 √ √ √ √ √ 2 2 4 3/2 8 4 4 2 −1 = 2 2− 2− 2 . =− = 2+ 2+ = 2 2− 3 3 3 3 3 3 1...
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