Thread: Help on these definite integrals

The answer: 3/16


Answer: 5/6

help

Originally Posted by ^_^Engineer_Adam^_^

The answer: 3/16

int{0:1}(z/(z^2 + 1)^3 dz

let u = z^2 + 1
=> du = 2z dz
=> (1/2) du = z dz

so our integral becomes:

(1/2)*int{u^-3}du
= (1/2)[(-1/2)u^-2]
= (1/2)[(-1/2)(z^2 + 1)^-2] evaluated between 0 and 1
= (1/2)[(-1/2)(2)^-2 - (-1/2)]
= 3/16

Originally Posted by ^_^Engineer_Adam^_^

Answer: 5/6

help

int{0:1}[(x^3 + 1)/(x + 1)]dx

let u = x + 1
=> du = dx

since u = x + 1
=> x = u - 1
=> x^3 = (u - 1)^3
=> x^3 + 1 = (u - 1)^3 + 1
=> x^3 + 1 = u^3 - 3u^2 + 3u - 1 + 1
=> => x^3 + 1 = u^3 - 3u^2 + 3u

so our integral becomes:

int{(u^3 - 3u^2 + 3u)/u}du
= int{u^2 - 3u + 3}du
= [(1/3)u^3 - (3/2)u^2 + 3u]

now we can write this in terms of x and evaluate, or we can change the limits in terms of u and plug them into the current solution, i won't do that though (it's more work), just letting you know you can. this is how if you're interested:

u = x + 1
if x = 0, u = 1
if x = 1, u = 2

so we can evaluate [(1/3)u^3 - (3/2)u^2 + 3u] between 1 and 2 as opposed to evaluating it between 0 and 1 with respect to x. but in terms of x, we have

[(1/3)(x + 1)^3 - (3/2)(x + 1)^2 + 3(x + 1)] evaluated between 0 and 1

= (1/3)2^3 - (3/2)2^2 + 3(2) - (1/3 - 3/2 + 3)
= 5/6

Originally Posted by ^_^Engineer_Adam^_^

Answer: 5/6

help

i just saw an even easier way

note that x^3 + y^3 = (x + y)(x^2 - xy + y^2)

we have in the numerator, x^3 + 1^3, we can write this therefore as (x + 1)(x^2 - x + 1) and the (x + 1) would cancel with the (x + 1) in the denominator. so our integral simply becomes:

int{x^2 - x + 1}dx
= [(1/3)x^3 - (1/2)x^2 + x] between 0 and 1
= 1/3 - 1/2 + 1
= 5/6

you can try to impress your professor with this way, even though it's not that complicated, its impressive that you see it