Integration By Substituition?

Question

Please help me to solve these problems:

1. ∫ from 0 to 2 of the function g(t) dt = 5. Calculate the following:
a.∫ from 0 to 4 of the function g(t/2) dt
b.∫ from 0 to 2 of the function g(2-t) dt

2. ∫ from 1 to 2 of sint/t dt

3. ∫ (x cos(x^2)/square root of (sin(x^2)) )dx

Answer 1

∫_0_2 g(t) dt = 5

(a)
y= ∫_0_4 g(t/2) dt
let t/2 =x ; so dt = 2dx; and the range becomes x:0 to 2
so, y= ∫_0_2 g(x) 2 dx = 2 ∫_0_2 g(x) dx = 2 * 5 = 10

(b)
y= ∫_0_2 g(2-t) dt
let 2-t=x; so dt = -dx; range becomes x:2 to 0
y= - - ∫_0_2 g(x) dx = 5

Answer 2

1. ∫ from 0 to 2 of g(t)dt = 5
If you ignore what g(t) actually is, and break this down you get:
G(2) - G(0) = 5

Now you can go on to the next ones.

a) ∫ from 0 to 4 of the function g(t/2) dt
Break this down too:
G(4/2) - G(0/2) = ?
Divide out the fractions and you get:
G(2) - G(0) = ?
From the information given you know that this equals 5.
G(2) - G(0) = 5

b) ∫ from 0 to 2 of the function g(2-t) dt
Break this down like in the first one
G(2-0) - G(2-2) = ?
G(2) - G(0) = ?
Now the two terms from the given info have just switched places. There is a rule that states if you flip the bounds (instead of going from 0 to 2 you go from 2 to 0) you just make the whole thing negative. So, use this here to get:
G(2) - G(0) = -5


2. ∫ from 1 to 2 of sint/t dt
Do you mean sin(t)/t ? I can't think of a way to integrate that...


3. ∫ (x cos(x^2)/square root of (sin(x^2)) )dx
Okay, make your u=sin(x^2). So:
u=sin(x^2)
du=xcos(x^2)dx
Substitute this in, now you have:
∫ (1/sqrt(u)) )dx
∫ (u^-1/2 )dx
-1/2(u^1/2)
-1/2(sqrt(u))
-1/2(sqrt(sinx^2))