function: x*cos(x^2+1)
with respect to x.
lower bound: 0
upper bound: 2
i got the integral of (1/2cosu)du from 0 to 2. do i need to do anything else or is my answer wrong?
with respect to x.
lower bound: 0
upper bound: 2
i got the integral of (1/2cosu)du from 0 to 2. do i need to do anything else or is my answer wrong?
-
You need to change your bounds
u = x^2 + 1
du = 2x * dx
x * cos(x^2 + 1) * dx =>
(1/2) * cos(u) * du
u = x^2 + 1
x = 0
u = 0 + 1 = 1
x = 2
u = 4 + 1 = 5
So the new integral is:
(1/2) * cos(u) * du
from
u = 1
to
u = 5
u = x^2 + 1
du = 2x * dx
x * cos(x^2 + 1) * dx =>
(1/2) * cos(u) * du
u = x^2 + 1
x = 0
u = 0 + 1 = 1
x = 2
u = 4 + 1 = 5
So the new integral is:
(1/2) * cos(u) * du
from
u = 1
to
u = 5
-
u = x^2 + 1, du = 2x dx, x dx = du/2
You have to change the limits of integration. When x = 0, u = 1 and when x = 2, u = 5 so after the substitution, the integral is from 1 to 5.
You have to change the limits of integration. When x = 0, u = 1 and when x = 2, u = 5 so after the substitution, the integral is from 1 to 5.