Question

Find the area of the region between y=x^6 and y=x^2 for 0 less than or equal to x less than or equal 1

Answer 1

on this interval \(x^2>x^6\) so your integral is
\[\int_0^1(x^2-x^6)dx\]

Answer 2

So I would take the integral of x squared minus integral of x to the sixth....im confused what do I plug in for x. 0 and 1?

Answer 3

find the anti-derivative using the power rule backwards
it is
\[\frac{x^3}{3}-\frac{x^7}{7}\] then replace x by 1

Answer 4

you get
\[\frac{1}{3}-\frac{1}{7}=\frac{4}{21}\]

Answer 5

you can skip evaluating at x = 0 because you will just get 0

Answer 6

ok

Answer 7

Thank You!

Answer 8

I have another problem but it is x^1/3 and x^1/5. How would I find the antiderivative of a x to a fraction?