- #1
coolguy1
- 1
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In the integral
integral(1,infinity) e^(-sqrt(x)) / sqrt(x)
STEP 1:
I let u = -sqrt(x)
du = -1/(2sqrt(x))
then my lower bound u = -1
then my upper bound u = -infinity
-2 integral(-1,infinity) e^u du
I would then switch the order of the integration bounds and multiply by -1My question is in the next problem integral(0,infinity) x^2/(1+x^3) dx
I let u = 1 + x^3
du = 3x^2 dx
du/3 = x^2 dx
lower bound u = 1
upper bound u = infinity
My question is: Would you multiply by -1 and switch the lower and upper bounds in this problem, or was that just the case in the previous problem?Thanks for your help and sorry I'm new to the commands and not sure how to use them yet.
integral(1,infinity) e^(-sqrt(x)) / sqrt(x)
STEP 1:
I let u = -sqrt(x)
du = -1/(2sqrt(x))
then my lower bound u = -1
then my upper bound u = -infinity
-2 integral(-1,infinity) e^u du
I would then switch the order of the integration bounds and multiply by -1My question is in the next problem integral(0,infinity) x^2/(1+x^3) dx
I let u = 1 + x^3
du = 3x^2 dx
du/3 = x^2 dx
lower bound u = 1
upper bound u = infinity
My question is: Would you multiply by -1 and switch the lower and upper bounds in this problem, or was that just the case in the previous problem?Thanks for your help and sorry I'm new to the commands and not sure how to use them yet.