Question

How to find this derivative of this integral?

Answer 1

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

Answer 2

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

Answer 3

Do you want the antiderivative?

Answer 4

I meant how to evaluate it, i read it wrong

Answer 5

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

Answer 6

How?

Answer 7

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

Answer 8

would you just bring the n down?

Answer 9

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

Answer 10

it would be one less than n?

Answer 11

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

Answer 12

it would just be n?

Answer 13

Nope.

Okay how about this...

What is the derivative of \(x^5\)?

Answer 14

\[5x ^{4}\]

Answer 15

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

Answer 16

nx? I dont know if i'm understanding?

Answer 17

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

Answer 18

What happens to the exponent?

Answer 19

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

Answer 20

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

Answer 21

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

Answer 22

n-1?

Answer 23

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

Answer 24

Yes, I understand that

Answer 25

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

Answer 26

what do we do?

Answer 27

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

Answer 28

okay

Answer 29

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

Answer 30

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: \[
b-1 = m\\ ab = 1
\]

Answer 31

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

Answer 32

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

Answer 33

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

Answer 34

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

Answer 35

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

Answer 36

x^2/2?

Answer 37

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

Answer 38

is it x^2?

Answer 39

Yes.

Answer 40

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 \(f(x)=x^2+C\).

Answer 41

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

Answer 42

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

Answer 43

a is -1 and b is 2

Answer 44

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

Answer 45

3

Answer 46

Good.

Answer 47

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

Answer 48

Thank you

Answer 49

Medal?

Answer 50

Yes, thank you