Evaluate the definite integral using substitution rule.

Question

Evaluate the definite integral using substitution rule.

1.integrate √(2x + 1) dx from 0 to 4

2.integrate x/(√(x + 1)) dx from 0 to 15

3.integrate x*(1 - x) ^ 3 dx from 0 to 0.5

4.integrate x *√(x - 3) dx from 3 to 7

5.integrate x * x^3 √(x + 1) dx from 0 to 7

Answer 1

1 sqr(2x+1) let u=2x+1 du/dx=2, dx =du/2

integral of (2x-1)dx= integral of (u/2)du = 2 times the integral of udu = 2u^2/2 = u^2 = (2x+1)^2

evaluated from 0 to 4 = (2(4)+1)^2 - 1 = 80

or evaluate u^2 from 1 to 9 = 81^2 -1^2 = 81-1 = 80

use the same u-substitution on the 2 through 5

2 x/sqr(x+1) = (u-1)/sqr(u) = u/sqru - 1/sqru = sqru - 1/sqru

integral of (u^(1/2) - u^(-1/2))dx = (2/3)u^(3/2) - (2)u^(1/2)

evaluated from 1 to 16 = (2/3)(16)^(3/2) - 2(16)^(1/2) -[2/3 -2] = 128/3 - 8 - 2/3 +2 = 126/3 -6 = 42-6 = 32

or integral of [(2/3)(x+1)^(3/2) - 2(x+1)^(1/2)]dx = (2/3)(x+1)^(3/2) - 2(x+1)^(1/2) evaluated from 0 to 15

= (2/3)(16)^(3/2) -2(16)^(1/2) - [(2/3) -2)]= (2/3)(64)-8 -2/3+2 = 126/3 -6 = 42-6 = 32

3 x(1-x)^3 = (1-u)u^3 = u^3 -u^4

integral of u^3 -u^4 = u^4/4 - u^5/5

evaluated from 0 to .5 = (.5)^4/4 - (.5)^5/5 = .0625/4 - .00625/5 = .015625 - .00125 = .014375

convert the x limits to u limits of evaluation

4 xsqr(x-3) = (u+3)sqr(u) = u^(3/2) + 3u^(1/2)

integral of (u+3)u^(1/2)dx = (2/5)u^(5/2) + 2u^(3/2)

evaluated from 0 to 4 = (2/5)(4)^(5/2) + 2(4)^(3/2) = (2/5)(32) + 2(8) = 64/5 + 16 = 12.8 +16 = 28.8

5 x(x^3)(sqr(x-1) = x^4(sqr(x-1) = (u+1)^4(sqr(u)) = (u^2 +2u +1)^2(sqru) = (u^4 + 4u^3 + 6u^2 + 4u + 1)u^(1/2) = u^(9/2) + 4u^(7/2) + 6u^(5/2) + 4u^(3/2) + u^(1/2)

integral = (2/11)u^(11/2) + (2/7)(4)u^(9/2) + (2/7)6u^(7/2) + (2/3)4u^(5/2) +2u^(3/2)

evaluated from u=1 to 8

= (2/11)(8)^(11/2) + (8/7)(8)^(9/2) + (12/7)(8^(7/2) + (8/3)(8)^(5/2) + 2(8)^(3/2) - 2/11-8/7-12/7-8/3 -2

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