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Evaluate the definite integral.
$ \displaystyle \int^2_1 x\sqrt{x - 1} \, dx $
$\frac{16}{15}$
01:27
Frank L.
Calculus 1 / AB
Chapter 5
Integrals
Section 5
The Substitution Rule
Integration
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Okay, we know Or you is X minus one. Which means our D'YOU is one D x, which means that now we know we have to alter our limits are lower limit is gonna be simply one minus one, which is zero, and an upper limit is gonna be to minus one, cause when exes to which gives us one. You did the 1/2 is the same thing, That squirt of you. Okay, now that we've simplified, we know we can take the integral use. The power will increase the exported by 1 to 5 over to divide by the new exponents. Now use the fundamental fear of calculus we can plug in. This is equivalent to 2/5 plus 2/3 minus zero, which some pause to 16 divided by 15.
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