Kayla Murray at 00:04, 27 October 2017

2017-10-27T00:04:00Z

Kayla Murray: Created page with "==Introduction== Let's say we want to integrate :: $\int x^2e^{x^3}~dx.$ Here, we can compute this antiderivative by using $..."$

2017-10-26T23:48:37Z

Created page with "==Introduction== Let's say we want to integrate ::<math>\int x^2e^{x^3}~dx.</math> Here, we can compute this antiderivative by using <math style="vertical-align: 0px">..."

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==Introduction==
Let's say we want to integrate

::<math>\int x^2e^{x^3}~dx.</math>

Here, we can compute this antiderivative by using  <math style="vertical-align: 0px">u-</math>substitution.

While  <math style="vertical-align: 0px">u-</math>substitution is an important integration technique, it will not help us evaluate all integrals.

For example, consider the integral

::<math>\int xe^x~dx.</math>

There is no substitution that will allow us to integrate this integral.

We need another integration technique called integration by parts.

The formula for integration by parts comes from the product rule for derivatives.

Recall from the product rule,

::<math>(f(x)g(x))'=f'(x)g(x)+f(x)g'(x).</math>

Then, we have

::<math>\begin{array}{rcl}
\displaystyle{f(x)g(x)} & = & \displaystyle{\int f'(x)g(x)+f(x)g'(x)~dx}\\
&&\\
& = & \displaystyle{\int f'(x)g(x)~dx+\int f(x)g'(x)~dx.}
\end{array}</math>

If we solve the last equation for the second integral, we obtain

::<math>\int f(x)g'(x)~dx = f(x)g(x)-\int f'(x)g(x)~dx.</math>

This formula is the formula for integration by parts.

But, as it is currently stated, it is long and hard to remember.

So, we make a substitution to obtain a nicer formula.

Let  <math style="vertical-align: -5px">u=f(x)</math>  and  <math style="vertical-align: -5px">dv=g'(x)~dx.</math>

Then,  <math style="vertical-align: -5px">du=f'(x)~dx</math>  and  <math style="vertical-align: -5px">v=g(x).</math>

Plugging these into our formula, we obtain

::<math>\int u~dv=uv-\int v~du.</math>

==Warm-Up==
Evaluate the following integrals.

'''1)'''   <math style="vertical-align: -7px">\int xe^x~dx</math>

{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
!Solution:  
|-
|Using the Product Rule, we have
|-
|
::<math>f'(x)=(x^2+x+1)(x^3+2x^2+4)'+(x^2+x+1)'(x^3+2x^2+4).</math>
|-
|Then, using the Power Rule, we have
|-
|
::<math>f'(x)=(x^2+x+1)(3x^2+4x)+(2x+1)(x^3+2x^2+4).</math>
|-
|-
|<u>NOTE:</u> It is not necessary to use the Product Rule to calculate the derivative of this function.
|-
|You can distribute the terms and then use the Power Rule.
|-
|In this case, we have
|-
|
::<math>\begin{array}{rcl}
\displaystyle{f(x)} & = & \displaystyle{(x^2+x+1)(x^3+2x^2+4)}\\
&&\\
& = & \displaystyle{x^2(x^3+2x^2+4)+x(x^3+2x^2+4)+1(x^3+2x^2+4)}\\
&&\\
& = & \displaystyle{x^5+2x^4+4x^2+x^4+2x^3+4x+x^3+2x^2+4} \\
&&\\
& = & \displaystyle{x^5+3x^4+3x^3+6x^2+4x+4.}
\end{array}</math>
|-
|Now, using the Power Rule, we get
|-
|
::<math>f'(x)=5x^4+12x^3+9x^2+12x+4.</math>
|-
|In general, calculating derivatives in this way is tedious. It would be better to use the Product Rule.
|}

{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|       <math>f'(x)=(x^2+x+1)(3x^2+4x)+(2x+1)(x^3+2x^2+4)</math>
|-
|or equivalently
|-
|       <math>f'(x)=x^5+3x^4+3x^3+6x^2+4x+4</math>
|}

'''2)'''   <math style="vertical-align: -14px">\int x\cos (2x)~dx</math>

{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
!Solution:  
|-
|
Using the Quotient Rule, we have
|-
|
::<math>f'(x)=\frac{x(x^2+x^3)'-(x^2+x^3)(x)'}{x^2}.</math>
|-
|Then, using the Power Rule, we have
|-
|
::<math>f'(x)=\frac{x(2x+3x^2)-(x^2+x^3)(1)}{x^2}.</math>
|-
|<u>NOTE:</u> It is not necessary to use the Quotient Rule to calculate the derivative of this function.
|-
|You can divide and then use the Power Rule.
|-
|In this case, we have
|-
|
::<math>\begin{array}{rcl}
\displaystyle{f(x)} & = & \displaystyle{\frac{x^2+x^3}{x}}\\
&&\\
& = & \displaystyle{\frac{x^2}{x}+\frac{x^3}{x}}\\
&&\\
& = & \displaystyle{x+x^2.} \\
\end{array}</math>
|-
|Now, using the Power Rule, we get
|-
|
::<math>f'(x)=1+2x.</math>
|}

{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
||       <math>f'(x)=\frac{x(2x+3x^2)-(x^2+x^3)}{x^2}</math>
|-
|or equivalently
|-
|       <math>f'(x)=1+2x</math>

|-
|}

'''3)'''   <math style="vertical-align: -14px">\int \ln x~dx</math>

{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
!Solution:  
|-
|Using the Quotient Rule, we get
|-
|
::<math>\begin{array}{rcl}
\displaystyle{f'(x)} & = & \displaystyle{\frac{\cos x(\sin x)'-\sin x (\cos x)'}{(\cos x)^2}}\\
&&\\
& = & \displaystyle{\frac{\cos x(\cos x)-\sin x (-\sin x)}{(\cos x)^2}}\\
&&\\
& = & \displaystyle{\frac{\cos^2 x+\sin^2 x}{\cos^2 x}} \\
&&\\
& = & \displaystyle{\frac{1}{\cos^2 x}}\\
&&\\
& = & \displaystyle{\sec^2 x}
\end{array}</math>
|-
|since  <math style="vertical-align: -2px">\sin^2 x+\cos^2 x=1</math>  and  <math style="vertical-align: -13px">\sec x=\frac{1}{\cos x}.</math>
|-
|Since  <math style="vertical-align: -14px">\frac{\sin x}{\cos x}=\tan x,</math>  we have
|-
|
::<math>\frac{d}{dx}{\tan x}=\sec^2 x.</math>
|}

{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|       <math>f'(x)=\sec^2 x</math>
|-
|}

== Exercise 1 ==

Evaluate  <math style="vertical-align: -13px">\int x^3 e^{-2x}~dx.</math>

First, we need to know the derivative of  <math style="vertical-align: 0px">\csc x.</math>  Recall

::<math>\csc x =\frac{1}{\sin x}.</math>

Now, using the Quotient Rule, we have

::<math>\begin{array}{rcl}
\displaystyle{\frac{d}{dx}(\csc x)} & = & \displaystyle{\frac{d}{dx}\bigg(\frac{1}{\sin x}\bigg)}\\
&&\\
& = & \displaystyle{\frac{\sin x (1)'-1(\sin x)'}{\sin^2 x}}\\
&&\\
& = & \displaystyle{\frac{\sin x (0)-\cos x}{\sin^2 x}}\\
&&\\
& = & \displaystyle{\frac{-\cos x}{\sin^2 x}} \\
&&\\
& = & \displaystyle{-\csc x \cot x.}
\end{array}</math>

Using the Product Rule and Power Rule, we have

::<math>\begin{array}{rcl}
\displaystyle{f'(x)} & = & \displaystyle{\frac{1}{x^2}(\csc x-4)'+\bigg(\frac{1}{x^2}\bigg)'(\csc x-4)}\\
&&\\
& = & \displaystyle{\frac{1}{x^2}(-\csc x \cot x+0)+(-2x^{-3})(\csc x-4)}\\
&&\\
& = & \displaystyle{\frac{-\csc x \cot x}{x^2}+\frac{-2(\csc x-4)}{x^3}.}
\end{array}</math>

So, we have
::<math>f'(x)=\frac{-\csc x \cot x}{x^2}+\frac{-2(\csc x-4)}{x^3}.</math>

== Exercise 2 ==

Evaluate  <math style="vertical-align: -5px">\int e^{3x}\sin (2x)~dx.</math>

Notice that the function  <math style="vertical-align: -5px">g(x)</math>  is the product of three functions.

We start by grouping two of the functions together. So, we have  <math style="vertical-align: -5px">g(x)=(2x\sin x)\sec x.</math>

Using the Product Rule, we get

::<math>\begin{array}{rcl}
\displaystyle{g'(x)} & = & \displaystyle{(2x\sin x)(\sec x)'+(2x\sin x)'\sec x}\\
&&\\
& = & \displaystyle{(2x\sin x)(\tan^2 x)+(2x\sin x)'\sec x.}
\end{array}</math>

Now, we need to use the Product Rule again. So,

::<math>\begin{array}{rcl}
\displaystyle{g'(x)} & = & \displaystyle{2x\sin x\tan^2 x+(2x(\sin x)'+(2x)'\sin x)\sec x}\\
&&\\
& = & \displaystyle{2x\sin x\tan^2 x+(2x\cos x+2\sin x)\sec x.}
\end{array}</math>

So, we have
::<math>g'(x)=2x\sin x\tan^2 x+(2x\cos x+2\sin x)\sec x.</math>

But, there is another way to do this problem. Notice

::<math>\begin{array}{rcl}
\displaystyle{g(x)} & = & \displaystyle{2x\sin x\sec x}\\
&&\\
& = & \displaystyle{2x\sin x\frac{1}{\cos x}}\\
&&\\
& = & \displaystyle{2x\tan x.}
\end{array}</math>

Now, you would only need to use the Product Rule once instead of twice.

== Exercise 3 ==

Evaluate  <math style="vertical-align: -16px">\int x\sec^2 x~dx.</math>

Using the Quotient Rule, we have

::<math>h'(x)=\frac{(x^2\cos x+3)(x^2\sin x+1)'-(x^2\sin x+1)(x^2\cos x+3)'}{(x^2\cos x+3)^2}.</math>

Now, we need to use the Product Rule. So, we have

::<math>\begin{array}{rcl}
\displaystyle{h'(x)} & = & \displaystyle{\frac{(x^2\cos x+3)(x^2(\sin x)'+(x^2)'\sin x)-(x^2\sin x+1)(x^2(\cos x)'+(x^2)'\cos x)}{(x^2\cos x+3)^2}}\\
&&\\
& = & \displaystyle{\frac{(x^2\cos x+3)(x^2\cos x+2x\sin x)-(x^2\sin x+1)(-x^2\sin x+2x\cos x)}{(x^2\cos x+3)^2}.}
\end{array}</math>

So, we get
::<math>h'(x)=\frac{(x^2\cos x+3)(x^2\cos x+2x\sin x)-(x^2\sin x+1)(-x^2\sin x+2x\cos x)}{(x^2\cos x+3)^2}.</math>

== Exercise 4 ==

Evaluate  <math style="vertical-align: -14px">\int x\sqrt{x+1}~dx.</math>

First, using the Quotient Rule, we have

::<math>\begin{array}{rcl}
\displaystyle{f'(x)} & = & \displaystyle{\frac{x^2\sin x (e^x)'-e^x(x^2\sin x)'}{(x^2\sin x)^2}}\\
&&\\
& = & \displaystyle{\frac{x^2\sin x e^x - e^x(x^2\sin x)'}{x^4\sin^2 x}.}
\end{array}</math>

Now, we need to use the Product Rule. So, we have

::<math>\begin{array}{rcl}
\displaystyle{f'(x)} & = & \displaystyle{\frac{x^2\sin x e^x - e^x(x^2(\sin x)'+(x^2)'\sin x)}{x^4\sin^2 x}}\\
&&\\
& = & \displaystyle{\frac{x^2\sin x e^x - e^x(x^2\cos x+2x\sin x)}{x^4\sin^2 x}.}
\end{array}</math>

So, we have
::<math>f'(x)=\frac{x^2\sin x e^x - e^x(x^2\cos x+2x\sin x)}{x^4\sin^2 x}.</math>

== Exercise 5 ==

Evaluate  <math style="vertical-align: -14px">\int \frac{\ln x}{x^3}~dx.</math>

First, using the Quotient Rule, we have

::<math>\begin{array}{rcl}
\displaystyle{f'(x)} & = & \displaystyle{\frac{x^2\sin x (e^x)'-e^x(x^2\sin x)'}{(x^2\sin x)^2}}\\
&&\\
& = & \displaystyle{\frac{x^2\sin x e^x - e^x(x^2\sin x)'}{x^4\sin^2 x}.}
\end{array}</math>

Now, we need to use the Product Rule. So, we have

::<math>\begin{array}{rcl}
\displaystyle{f'(x)} & = & \displaystyle{\frac{x^2\sin x e^x - e^x(x^2(\sin x)'+(x^2)'\sin x)}{x^4\sin^2 x}}\\
&&\\
& = & \displaystyle{\frac{x^2\sin x e^x - e^x(x^2\cos x+2x\sin x)}{x^4\sin^2 x}.}
\end{array}</math>

So, we have
::<math>f'(x)=\frac{x^2\sin x e^x - e^x(x^2\cos x+2x\sin x)}{x^4\sin^2 x}.</math>

== Exercise 6 ==

Evaluate  <math style="vertical-align: -14px">\int \sin(2x)\cos(3x)~dx.</math>

First, using the Quotient Rule, we have

::<math>\begin{array}{rcl}
\displaystyle{f'(x)} & = & \displaystyle{\frac{x^2\sin x (e^x)'-e^x(x^2\sin x)'}{(x^2\sin x)^2}}\\
&&\\
& = & \displaystyle{\frac{x^2\sin x e^x - e^x(x^2\sin x)'}{x^4\sin^2 x}.}
\end{array}</math>

Now, we need to use the Product Rule. So, we have

::<math>\begin{array}{rcl}
\displaystyle{f'(x)} & = & \displaystyle{\frac{x^2\sin x e^x - e^x(x^2(\sin x)'+(x^2)'\sin x)}{x^4\sin^2 x}}\\
&&\\
& = & \displaystyle{\frac{x^2\sin x e^x - e^x(x^2\cos x+2x\sin x)}{x^4\sin^2 x}.}
\end{array}</math>

So, we have
::<math>f'(x)=\frac{x^2\sin x e^x - e^x(x^2\cos x+2x\sin x)}{x^4\sin^2 x}.</math>

← Older revision		Revision as of 00:04, 27 October 2017
Line 1:		Line 1:
	==Introduction==		==Introduction==
		+	The derivative of the function  <math>f(x)</math>  is the instantaneous rate of change of  <math>f(x).</math>
		+
		+	Let  <math>h</math> be a nonzero number.
		+
		+	The average rate of change of the function  <math>f(x)</math>  on the interval
		+
	Let's say we want to integrate		Let's say we want to integrate

Limit Definition of Derivative - Revision history

Kayla Murray at 00:04, 27 October 2017

Kayla Murray: Created page with "==Introduction== Let's say we want to integrate ::\int x^2e^{x^3}~dx. Here, we can compute this antiderivative by using ..."

Kayla Murray: Created page with "==Introduction== Let's say we want to integrate :: $\int x^2e^{x^3}~dx.$ Here, we can compute this antiderivative by using $..."$