# Math Help - Integration :(

1. ## Integration :(

Im doing q's were i need to find the area of curves on a graph but im going wrong somewhere. please help

y=x^5-2 between x=-1 and x=0 if differentiated to 1/6x^6 - 2x
and drawn the graph of x^5-2

then by substituting x to be o and -1
{1/6(-1)^6 - 2(-1)}-{1/6(0)^6-2(0)]
and got the answer to be25/6 but the answer in the book is 2 1/6
thank you

2. Originally Posted by Chez_
Im doing q's were i need to find the area of curves on a graph but im going wrong somewhere. please help

y=x^5-2 between x=-1 and x=0 if differentiated to 1/6x^6 - 2x
and drawn the graph of x^5-2

then by substituting x to be o and -1
{1/6(-1)^6 - 2(-1)}-{1/6(0)^6-2(0)]
and got the answer to be25/6 but the answer in the book is 2 1/6
thank you

Your post is a bit muddled but all you need to do is work out

$\frac{1}{6}(-1)^6-2\times-1$