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Evaluate the integral.
$ \displaystyle \int^8_1 x^{-2/3} \,dx $
$$\int_{1}^{8} x^{-2 / 3} d x=3$$
00:30
Frank L.
Calculus 1 / AB
Chapter 5
Integrals
Section 3
The Fundamental Theorem of Calculus
Integration
Campbell University
Harvey Mudd College
Baylor University
Boston College
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integrate meaning. Add one to the exponents and divide by the new exponents. We can simplify this to be three x to the 1/3 from the bounds of 1 to 8. This is three times two months, three times one, which is six months three, which is three.
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