Evaluate the following integral: ∫ 2 3 x 5 2 d x \int_2^3\!x^{\frac52}\,dx ∫ 2 3 x 2 5 d x

this problem, we are asked to evaluate the following integral, and we are given the integral from two to three of x to the five halves power. Now we can solve this by using the fundamental theorem of calculus part two, which which tells us that we can just take the antiderivative of this portion and then evaluate the bounds. Now first take the antiderivative of x to the five halves power, and that's gonna give me x to the five halves power plus one over five halves plus one. And if we're doing an indefinite integral, we'd have to do a plus c. But since we have a definite integral here, we don't need to worry about the constant at the end. We just need to make sure we bound this from the two upper and lower bounds that we have, respectively. So the upper and lower bounds here. And now let's go ahead and figure out what these fractions are simplified. So we have five to have five halves plus one, which is the same thing as x to the seven halves. Right? That's gonna be what we have on top, and that's gonna be divided by seven halves on bottom, all bounded from two to three. Then we can go ahead and simplify this even farther. By recognizing the seven halves, I can just flip and bring up to the top here, which can give me two times x to the seventh halves power over seven, and that's gonna be evaluated from two to three. Now from here, what I need to do is plug in my bounds respectively. So we're gonna start by plugging in the high bound of three. So it's gonna give me two times, and then we're gonna have two times three to the seven halves power all divided by seven. And then what I'll do is subtract off my lower bound plugged in. So we're going to have this two right here. So we have two times two to the seven halves power, all divided by seven. And at this point, we got this kind of mess of fractions right here. I think the best thing to do here is gonna be to evaluate all this on a calculator. And if we do that, you should get an approximation of 10.129 as your result. So that would be the solution, the approximate solution for what we have for this definite integral, and that is how you can solve this problem. So hope you found this video helpful, and let's move on.

8. Definite Integrals

Fundamental Theorem of Calculus

Business Calculus

8. Definite Integrals

Fundamental Theorem of Calculus

Struggling with Business Calculus?

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Multiple Choice

Evaluate the following integral:
$\int_2^3\!x^{\frac52}\,dx$

19.360

16.594

10.129

11.817

Verified step by step guidance

Rewrite the integral in a more readable form: $ \int_2^3 x^{\frac{5}{2}} \, dx $. This represents the definite integral of $ x^{\frac{5}{2}} $ from 2 to 3.

Apply the power rule for integration: $ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C $, where $ n \neq -1 $. Here, $ n = \frac{5}{2} $.

Using the power rule, the antiderivative of $ x^{\frac{5}{2}} $ is $ \frac{x^{\frac{5}{2} + 1}}{\frac{5}{2} + 1} = \frac{x^{\frac{7}{2}}}{\frac{7}{2}} = \frac{2}{7} x^{\frac{7}{2}} $.

Evaluate the definite integral by substituting the upper and lower limits (3 and 2) into the antiderivative: $ \left[ \frac{2}{7} x^{\frac{7}{2}} \right]_2^3 = \frac{2}{7} (3^{\frac{7}{2}}) - \frac{2}{7} (2^{\frac{7}{2}}) $.

Simplify the expression $ \frac{2}{7} (3^{\frac{7}{2}}) - \frac{2}{7} (2^{\frac{7}{2}}) $ to find the final value of the definite integral.