Fundemental Theorem Questions (I think)

**skirk34** · August 1st 2008, 03:58 AM

For maths, i dont know how to work out the following:

1)Find the value of;
Integral(0,4): f'(x) ( (f(x))^3 -2f(x) -3) dx
given that f(0) = 2, f'(0) = -6, f(4) = 3 and f'(4) = -3

2) y= Integral(5,5x^4):sin( t^2 - 5 ) dt
whats y'?

(I'm new so sorry, i dont know how to write in equation form)

**wingless** · August 1st 2008, 04:19 AM

Hint:

$\int f'(x) [f(x)]^n~dx$

Let $u = f(x)$ , then $du = f'(x)~dx$ .

$\int u^n~du$

- If $n\neq -1$ , then it's $\frac{u^{n+1}}{n+1}= \frac{[f(x)]^{n+1}}{n+1}$ .

- If $n = -1$ , then it's $\ln u = \ln f(x)$

**kalagota** · August 1st 2008, 04:24 AM

Originally Posted by skirk34

For maths, i dont know how to work out the following:

1)Find the value of;
Integral(0,4): f'(x) ( (f(x))^3 -2f(x) -3) dx
given that f(0) = 2, f'(0) = -6, f(4) = 3 and f'(4) = -3

set $u = f(x)$
$du = f'(x)dx$

at $x=0: u=f(0)=2$
at $x=4: u=f(4)=3$

thus

$\int_2^3 (u^3 -2u -3)du = ...$

**kalagota** · August 1st 2008, 04:28 AM

Originally Posted by skirk34

For maths, i dont know how to work out the following:

2) y= Integral(5,5x^4):sin( t^2 - 5 ) dt
whats y'?

(I'm new so sorry, i dont know how to write in equation form)

if $F(x) = \int_a^{f(x)} h(t)dt$ where $a$ is a constant,

then $F'(x) = h(f(x)) \, f'(x)$

**skirk34** · August 1st 2008, 05:04 AM

For the second question I got the answer:
y'(x) = sin(x^2 - 5).20x^3

The quiz its from isn't liking my answer though

Heres the hint it gives me:
Let the integrand be f(t) and let an antiderivative of f(t) be F(t). Then y(x) = F(5 x^4) - F(5), where F(5) is just a constant.

**kalagota** · August 1st 2008, 05:09 AM

Originally Posted by skirk34

For the second question I got the answer:
y'(x) = sin(x^2 - 5).20x^3

The quiz its from isn't liking my answer though

Heres the hint it gives me:
Let the integrand be f(t) and let an antiderivative of f(t) be F(t). Then y(x) = F(5 x^4) - F(5), where F(5) is just a constant.

im sorry.. i edited my last post.. take a look at it..

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