How to find this derivative of this integral?

- anonymous

How to find this derivative of this integral?

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- anonymous

\[\int\limits_{-1}^{2} 2xdx\]

- anonymous

\[
\frac{d}{dx}\int^{g(x)}_af(t)dt = f(g(x))g'(x)
\]

- anonymous

Do you want the antiderivative?

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## More answers

- anonymous

I meant how to evaluate it, i read it wrong

- anonymous

Okay first you need to find the anti-derivative. Do you know how to do that?

- anonymous

How?

- anonymous

First, do you know how to find the derivative of the following function: \[
\frac{d}{dx} (x^n)
\]

- anonymous

would you just bring the n down?

- anonymous

Yeah, what is the derivative in terms of \(n\)?

- anonymous

it would be one less than n?

- anonymous

One less than \(n\)? Are you saying \(n-1\)?
What is the derivative of \(x^n\)?

- anonymous

it would just be n?

- anonymous

Nope.
Okay how about this...
What is the derivative of \(x^5\)?

- anonymous

\[5x ^{4}\]

- anonymous

Okay, so suppose \(5=n\). What is the derivative of \(x^n\)?

- anonymous

nx? I dont know if i'm understanding?

- anonymous

\[
x^5 \to 5x
\]What happened to the \(4\)?

- anonymous

What happens to the exponent?

- anonymous

I'm trying to get you to understand the general concept, then you'll be able to do many problems on your own.

- anonymous

What happens to the 4? isn't it supposed to be the new exponent?

- anonymous

Yes. So what is the new exponent when you take the derivative of \(x^n\)?

- anonymous

n-1?

- anonymous

\[
\frac{d}{dx}x^n = nx^{n-1}
\]

- anonymous

Yes, I understand that

- anonymous

Okay, now suppose we want to reverse it...

- anonymous

what do we do?

- anonymous

So suppose we want to find the anti derivative of \(x^m\)

- anonymous

okay

- anonymous

We need to find \(a\) and \(b\):\[
\frac{d}{dx} ax^b = x^m
\]

- anonymous

We know that \[
\frac{d}{dx}ax^b = abx^{b-1}
\]So \[
\color{red}{ab}x^{\color{blue}{b-1}} = x^{\color{blue}{m}}
\]This means two things: \[
b-1 = m\\ ab = 1
\]

- anonymous

We can see that \[
b = m+1
\]So we have found \(b\). But what about \(a\)?\[
a = \frac 1 b = \frac 1 {m+1}
\]

- anonymous

So \[
\frac{d}{dx}ax^b = \frac{d}{dx} \frac{1}{m+1} x^{m+1} = x^m
\]

- anonymous

In sort, the antiderivative of \(x^m\) is \[
\frac{x^{m+1}}{m+1}
\]

- anonymous

okay, so how do i evaluate the integral....?

- anonymous

What is the antiderivative of \(x^1\)?

- anonymous

x^2/2?

- anonymous

Yes. What is the anti derivative of \(2x\) then?

- anonymous

is it x^2?

- anonymous

Yes.

- anonymous

Technically you need a constant of integration so it is \(x^2+C\)
Anyway\[
\int^b_af'(x)dx = f(b)-f(a)
\]In this case \(f'(x)=2x\) and we have found that \(f(x)=x^2+C\).

- anonymous

The \(C\) will cancel out when we subtract, so we don't need to worry about it.

- anonymous

So what is \(b\) and what is \(a\) in this case?

- anonymous

a is -1 and b is 2

- anonymous

Okay so what is \(f(b)-f(a)\) in this case?

- anonymous

3

- anonymous

Good.

- anonymous

\[
\int^{2}_{-1}2x\;dx = (2)^2-(-1)^2=3
\]

- anonymous

Thank you

- anonymous

Medal?

- anonymous

Yes, thank you

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