Solve my Calculus 1 problem. Here is a photo.

Answered question

2022-03-22

Answer & Explanation

karton

Expert2023-04-25Added 613 answers

To evaluate the integral $\int_{0}^{1} \int_{\sqrt{y}}^{1} \sqrt{x^{3} + 1} d x d y$ by reversing the order of integration, we first draw the region of integration in the xy-plane.

The region is bounded by the lines x = 0, x = 1, and y = x^2. We can see that x ranges from 0 to 1 and y ranges from 0 to 1.

Thus, the integral can be written as $\int_{0}^{1} \int_{0}^{x^{2}} \sqrt{x^{3} + 1} d y d x$ .

To reverse the order of integration, we need to rewrite the limits of integration for x and y. Since y ranges from 0 to $x^{2}$ , we can write x as a function of y: $x = \sqrt{y}$ .

Thus, the integral becomes $\int_{0}^{1} \int_{0}^{1} \sqrt{x^{3} + 1} d x d y = \int_{0}^{1} \int_{0}^{y^{2}} \sqrt{x^{3} + 1} d x d y$ .

Now, we can evaluate the inner integral with respect to x:

$\int_{0}^{y^{2}} \sqrt{x^{3} + 1} d x = (\frac{1}{2}) \frac{{(x^{3} + 1)}^{\frac{3}{2}}}{\frac{3}{2}}] ∣_{0}^{y^{2}}$
$= (\frac{1}{3}) \frac{{(y^{6} + 1)}^{\frac{3}{2}} - 2}{2}$

Substituting this result back into the original integral, we have:

$\int_{0}^{1} \int_{0}^{1} \sqrt{x^{3} + 1} d x d y = \int_{0}^{1} [(\frac{1}{3}) \frac{{(y^{6} + 1)}^{\frac{3}{2}} - 2}{2}] d y$
$= (\frac{1}{6}) [(2 \sqrt{2} - 2) + \frac{2 \sqrt{2} - 2}{7}]$

Therefore, the value of the integral is $(\frac{1}{21}) (4 \sqrt{2} - 4)$ .

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Calculus 1

Didn't find what you were looking for?