# Math Help - Evaluate the intergral

1. ## Evaluate the intergral

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2. Originally Posted by dizizviet

2.

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

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

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

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.

3. can anyone else help me figure out the last three equations?

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

5. 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 =)