Some integral problems I'm having trouble with

• Feb 28th 2007, 06:08 PM
coolio
Some integral problems I'm having trouble with
1. The integral: (sinˆ2 (3x) cos(3x) ) dx

2. The integral: (x+2)(x-3)ˆ20 dx

3. The integral from 0 to pi/4: sqrt(tan(x)) secˆ2 (x) ) dx

4. The integral from 0 to 1: ( eˆ(cubed root of x) ) / 6xˆ(2/3) dx

5. Find the area that is enclosed by the curve y=2x-xˆ3 , the line x=1, the line x=5, and the x axis.
• Feb 28th 2007, 07:49 PM
coolio
Integral problems!
1. The integral: (x+2)(x-3)ˆ20 dx

2. The integral from 0 to 1: ( eˆ(cubed root of x) ) / 6xˆ(2/3) dx

3. The integral from 0 to pi/4: sqrt(tan(x)) secˆ2 (x) ) dx

4. The integral: (sinˆ2 (3x) cos(3x) ) dx

5. Find the area that is enclosed by the curve y=2x-xˆ3 , the line x=1, the line x=5, and the x axis.
• Feb 28th 2007, 09:11 PM
CaptainBlack
Quote:

Originally Posted by coolio
1. The integral: (x+2)(x-3)ˆ20 dx

The integrand is the derivative of f(x)=(ax+b)(x-3)^21 for same a and b.
Differentiating f(x) gives:

f'(x) = (x-3)^20 [22*a*x + 21*(ax+b)]

equating coefficients gives a=1/22, and b=47/(21*22)

So by the fundamental theorem of calculus:

Integral (x+2)(x-3)ˆ20 dx = (1/22)(x+47/21)(x-3)^21 +C

RonL
• Feb 28th 2007, 09:27 PM
earboth
first 2 problems only
Quote:

Originally Posted by coolio
1. The integral: (sinˆ2 (3x) cos(3x) ) dx

2. The integral: (x+2)(x-3)ˆ20 dx

3. The integral from 0 to pi/4: sqrt(tan(x)) secˆ2 (x) ) dx

4. The integral from 0 to 1: ( eˆ(cubed root of x) ) / 6xˆ(2/3) dx

5. Find the area that is enclosed by the curve y=2x-xˆ3 , the line x=1, the line x=5, and the x axis.

Hi,

to 1.) Use substitution:

u(x) = sin(3x) ===> du/dx = 3*cos(3x) ===> du = 3*cos(3x) * dx

(1/3)∫(sin²(3x) * 3*cos(3x) dx = (1/3) ∫u²*du = 1/9*u³ + C

resubstitute:

∫(sin²(3x) * cos(3x) dx = 1/9*sin³(3x) + C

to 2.)
Use partial integration:

Code:

```∫(x+2)(x-3)^20 dx = 1/21*(x+2)(x-3)^21 - ∫1*1/21*(x-3)^21 dx u  = x+2      v  = 1/21*(x-3)^21 u' = 1        v' = (x-3)^20```

∫(x+2)(x-3)^20 dx = 1/21*(x+2)(x-3)^21 - 1/462*(x-3)^22 + C

EB
• Mar 1st 2007, 12:23 AM
earboth
Quote:

Originally Posted by coolio
...
5. Find the area that is enclosed by the curve y=2x-xˆ3 , the line x=1, the line x=5, and the x axis.

Hi,

in the given interval there is a zero of the function. Therefore you have to split the interval into [1, √(2)] and ]√(2), 5].

∫ from 1 to √(2) (2x-x³)dx = (x² - 1/4x^4) from 1 to √(2) = (2-1) - (1-1/4) = 1/4

∫ from √(2) to 5 (2x-x³)dx = (x² - 1/4x^4) from √(2) to 5 = (22-625/4) - (2-1) = -529/4

Add the absolute values. The area is 1/4 + 529/4 = 530/4 = 265/2 = 1325.5

EB
• Mar 1st 2007, 05:43 AM
Soroban
Hello, coolio!

Quote:

3) .(tan x)^½ sec²x dx
Let: u = tan x

Quote:

. . . . .e^(x^{1/3})
4) .
--------------- dx
. . . . . .6x^{2/3}

Let: u = x^{1/3}