# Evaluate the intergral

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• May 11th 2009, 03:57 PM
dizizviet
Evaluate the intergral
• May 11th 2009, 05:39 PM
pickslides
Quote:

Originally Posted by dizizviet

$\int_1^2 9x^2\times 2x^3 dx$

$\int_1^2 9\times2x^{2+3} dx$

$\int_1^2 18x^{5} dx$

now use the following formula

$\int ax^{n} dx = \frac{ax^{n+1}}{n+1}+c$

to find

$\int_1^2 18x^{5} dx = \frac{18x^{5+1}}{5+1}$

$= \frac{18x^{6}}{6}$

$= 3x^{6}$

We we are intergrating between 1 and 2 sub these values in and subtract the values

$= 3(2)^{6} - 3(1)^{6}$

$= 3\times 32 - 3\times 1$

$= 96- 3 = 93$

this will work for Q4.
• May 13th 2009, 06:09 AM
dizizviet
can anyone else help me figure out the last three equations? (Rofl)
• May 13th 2009, 06:15 AM
mr fantastic
Quote:

Originally Posted by dizizviet
can anyone else help me figure out the last three equations? (Rofl)

Post #2 showed you how to do Q2 and told you that Q4 was done in the same way.

Q1 Substitute w = 7t and recognise a standard form involving a hyperbolic function.

Q3 Substitute w = log(9x).
• May 13th 2009, 11:35 AM
dizizviet
Quote:

Originally Posted by pickslides
$\int_1^2 9x^2\times 2x^3 dx$

$\int_1^2 9\times2x^{2+3} dx$

$\int_1^2 18x^{5} dx$

now use the following formula

$\int ax^{n} dx = \frac{ax^{n+1}}{n+1}+c$

to find

$\int_1^2 18x^{5} dx = \frac{18x^{5+1}}{5+1}$

$= \frac{18x^{6}}{6}$

$= 3x^{6}$

We we are intergrating between 1 and 2 sub these values in and subtract the values

$= 3(2)^{6} - 3(1)^{6}$

$= 3\times 32 - 3\times 1$

$= 96- 3 = 93$

this will work for Q4.

I'm sorry i missed typed my equations for the the second problem and third problem
but i firgured it out though thanks alot for your helps =)