Originally Posted by
pickslides $\displaystyle \int_1^2 9x^2\times 2x^3 dx$
$\displaystyle \int_1^2 9\times2x^{2+3} dx$
$\displaystyle \int_1^2 18x^{5} dx$
now use the following formula
$\displaystyle \int ax^{n} dx = \frac{ax^{n+1}}{n+1}+c$
to find
$\displaystyle \int_1^2 18x^{5} dx = \frac{18x^{5+1}}{5+1}$
$\displaystyle = \frac{18x^{6}}{6}$
$\displaystyle = 3x^{6}$
We we are intergrating between 1 and 2 sub these values in and subtract the values
$\displaystyle = 3(2)^{6} - 3(1)^{6}$
$\displaystyle = 3\times 32 - 3\times 1$
$\displaystyle = 96- 3 = 93$
this will work for Q4.