# Definite integration help.

• Jun 7th 2009, 09:48 AM
pantsaregood
Definite integration help.
My logic may be horrible.

Differentiate g(x)= definite integral on [2, x^2+3x-1] sec(t)dt and then prove the theorem you applied.

= antiderivative of sec(x^2+3x-1) - antiderivative of sec(2)

d/dx(ntiderivative of sec(x^2+3x-1) - antiderivative of sec(2))

= sec(x^2+3x-1) - sec(2)

Which is exactly what the integral originally states, isn't it?
• Jun 7th 2009, 11:06 AM
TheEmptySet
Quote:

Originally Posted by pantsaregood
My logic may be horrible.

Differentiate g(x)= definite integral on [2, x^2+3x-1] sec(t)dt and then prove the theorem you applied.

= antiderivative of sec(x^2+3x-1) - antiderivative of sec(2)

d/dx(ntiderivative of sec(x^2+3x-1) - antiderivative of sec(2))

= sec(x^2+3x-1) - sec(2)

Which is exactly what the integral originally states, isn't it?

If I understand this correctly we have

$\displaystyle g(x)=\int_{2}^{x^2+3x-1}\sec(t)dt$

This can be viewed as a function composition. Let $\displaystyle u=x^2+3x-1$

Then $\displaystyle g(u)=\int_{2}^{u}\sec(t)dt$

Now by the chain rule

$\displaystyle \frac{d}{dx} g(x)=\frac{d}{dx}g(u)=g'(u)\cdot u'=g'(x^2+3x-1)(2x+3)$

Remember that the fundemental theorem of calculus says that(Notice the upper limit of integration) check you book for the hypothsis

$\displaystyle \frac{d}{dx} \int _{c}^{x}f(t)dt=f(x)$

Combining these two results we get

$\displaystyle \frac{d}{dx} g(x)=\frac{d}{dx}g(u)=g'(u)\cdot u'=g'(x^2+3x-1)(2x+3)=(2x+3)\sec(x^2+3x-1)$

Notice that if you integrate both sides of the last equaion with the same u substition that we did above you will get the same function g back. i.e $\displaystyle u=x^2+3x-1$

I hope this helps