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

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

anonymous · Answer

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

anonymous · Answer

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?

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

ok

anonymous · Answer

Thank You!

anonymous · Answer

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?