Workshop_Solutions_5

# 2 4 mathematical methods for physics 1 solutions to

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

Ask a homework question - tutors are online