# Math Help - Integrals: #1 Help with fraction #2 Moment of inertia

1. ## Integrals: #1 Help with fraction #2 Moment of inertia

Problems:
25-2-EX9
The time rate of change of the displacement (velocity) of a robot arm is ds/dt = 8t/(t^2 + 4)^2. Find the expression for the displacement as a function of time if s = -1 m when t = 0 s.
26-5-9
Find the moment of inertia of a plate covering the first-quadrant region bounded by y^2 = x, x = 9 and the x-axis with respect to x-axis.

Attempt
25-2-EX9
This is a book example, so the solution is here. I am hung up on one step though.
ds = (int)/(8tdt)((t^2 + 4)^2) = 4 (int) (t^2 + 4)^-2(2t dt)

This next step throws me off. What happened to (2t dt)??

it just disappears. Can someone explain? Thanks.

s = 4(1)(-1))(t^2 + 4)^-1 + C

26-5-9
Ix =k (int) y^2(x2 - x1)dy

9ky^2 dx = 9kx = 9k1/2x^2

4.5x^2 = 4.5 * 3^2

4.5*9 = 40.5

However, the answer is 162/5k = 32.4

so where'd I go wrong? thanks.

2. ## Re: Integrals: #1 Help with fraction #2 Moment of inertia

Originally Posted by togo
...

Attempt
25-2-EX9
This is a book example, so the solution is here. I am hung up on one step though.
ds = (int)/(8tdt)((t^2 + 4)^2) = 4 (int) (t^2 + 4)^-2(2t dt)

This next step throws me off. What happened to (2t dt)??

it just disappears. Can someone explain? Thanks.

...
If you have
$\int(t^2+4)^{-2} \cdot 2t dt$
Then use integration by substitution:

$u = t^2+4~\implies~\frac{du}{dt} = 2t~\implies~du = 2t dt$

$\int u^{-2} du = -u^{-1}+c$
$\int(t^2+4)^{-2} \cdot 2t dt = -(t^2+4)^{-1}+c$