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# Lost with integration

[From: ] [author: ] [Date: 11-05-20] [Hit: ]
Can someone please help me? Im stuck on a couple of them.The definite integral from 0 to pi (upper bound= pi,The definite integral from 4 to 1 (upper bound= 4,Thank you so much! Im just really lost!......
In my calculus class, we're working on integration by parts specifically, but the worksheet I was given can utilize various methods. Can someone please help me? I'm stuck on a couple of them.

The definite integral from 0 to pi (upper bound= pi, lower bound =0)
of tsin3t dt

The definite integral from 4 to 1 (upper bound= 4, lower bound= 1)
of [sqrt(t)]lnt dt

Thank you so much! I'm just really lost! ><

-
t * sin(3t) * dt
u = t
du = dt
dv = sin(3t) * dt
v = (-1/3) * cos(3t)

int(v * du) =
uv - int(u * dv) =
(-1/3) * t * cos(3t) + (1/3) * int(cos(3t) * dt) =>
(-1/3) * t * cos(3t) + (1/3) * (1/3) * sin(3t) + C =>
(1/9) * sin(3t) - (1/3) * t * cos(3t) + C

From 0 to pi

(1/9) * sin(3pi) - (1/3) * pi * cos(3pi) - (1/9) * sin(0) + (1/3) * 0 * cos(0) =>
(1/9) * 0 - (1/3) * pi * 1 - (1/9) * 0 + 0 =>
(-1/3) * pi

sqrt(t) * ln(t) * dt
u = ln(t)
du = dt / t
dv = t^(1/2) * dt
v = (2/3) * t^(3/2)

uv - int(v * du) =>
ln(t) * (2/3) * t^(3/2) - (2/3) * int(t^(3/2) * dt / t) =>
(2/3) * t * sqrt(t) * ln(t) - (2/3) * int(t^(1/2) * dt) =>
(2/3) * t * sqrt(t) * ln(t) - (2/3) * (2/3) * t^(3/2) + C =>
(2/3) * t * sqrt(t) * (ln(t) - (2/3)) + C

from 1 to 4

(2/3) * 4 * sqrt(4) * (ln(4) - (2/3)) - (2/3) * 1 * sqrt(1) * (ln(1) - (2/3)) =>
(8/3) * 2 * (ln(4) - (2/3)) - (2/3) * (0 - (2/3)) =>
(16/3) * (ln(4) - (2/3)) + (4/9) =>
(4/3) * (4 * (ln(4) - (2/3)) + (1/3)) =>
(4/3) * (4ln(4) - (8/3) + (1/3)) =>
(4/3) * (12 * ln(4) / 3 - 7/3) =>
(4/9) * (12 * 2 * ln(2) - 7) =>
(4/9) * (24 * ln(2) - 7)
1
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