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Integrate each of the given functions.$$\int_{2.5 y \sqrt{4 y^{2}-9}}^{3} \frac{d y}{\sqrt{4 y^{2}-9}}$$
Calculus 1 / AB
Chapter 28
Methods of Integration
Section 8
Integration by Trigonometric Substitution
Integrals
Campbell University
Oregon State University
Baylor University
University of Nottingham
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the integral of ex route X squared plus three. Why d y from 4 to 5? Uh, note that the d Y here means that we can treat all other variables as constants. So this X first of all, can come out of the integral. And we'll rewrite X squared plus three y to the power of 1/2. Using our exponents laws, the anti derivative for X squared plus three y to the power of 1/2 is to over three times x squared plus three y to the power of three over to make sure you account for the chain rule with the 1/3 at the end. And we evaluate this from 4 to 5 well below the constants which will make this to X over nine times. Ah, the substitution of the Y values. So X squared plus 15 to the power of three over two, minus x squared. Plus 12th to the power of three over two. Yeah, I don't know. Does that sound okay?
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