Find the integral of

x times the square root of (400-100x^2)dx

intgrl(x(400-100x^2)^(1/2) dx

= (2/3)x(400-100x^2)^(3/2) * 1/(-200x) + c
= (-1/300) (400 - 100x^2)^(3/2) + c

Thank you!

To find the integral of \(x\) multiplied by the square root of \(400-100x^2\), we can use a u-substitution.

Let's start with the substitution:
\(u = 400-100x^2\)

First, calculate \(du/dx\):
\(du/dx = -200x\)

To solve for dx, we can rearrange this equation:
\(dx = du/(-200x)\)

Now, substitute the value of \(u\) and \(dx\) in the integral:
\(\int (x \sqrt{400-100x^2})dx = \int \left(\frac{u}{(-200x)}\right)dx\)

Now, let's simplify the expression further:
\(\int \left(\frac{u}{(-200x)}\right)dx = \int \left(\frac{u}{(-200x)}\right)\left(\frac{-1}{200}\right)\left(\frac{-200}{-1}\right) dx\)
\(\int \left(\frac{-u}{200x}\right)dx = \int \left(\frac{-u}{200x}\right) \left(\frac{-200}{-1}\right) dx \)
\(\int \frac{u}{x} dx = \int \frac{u}{x} \cdot (-1) dx\)
\(-\int \frac{u}{x} dx \)

Now, we can integrate with respect to u:
\(-\int \frac{u}{x} dx = -\int \frac{u}{x} du\)

Now, let's simplify the expression further:
\(-\frac{1}{2} \int \frac{u}{x} du = -\frac{1}{2} \int \frac{u}{x} \frac{dx}{dx} du\)
\(-\frac{1}{2} \int \frac{u}{x} \frac{dx}{dx} du = -\frac{1}{2} \int \frac{u}{x} \frac{d(xu)}{dx} du\)

Using the product rule, we can simplify it even more:
\(-\frac{1}{2} \int \frac{u}{x} \frac{d(xu)}{dx} du = -\frac{1}{2} \int d(xu) du\)

Integrating \(d(xu)\) with respect to u:
\(-\frac{1}{2} \int d(xu) du = -\frac{1}{2} \int du(xu)\)

Finally, substitute the value of u back in terms of x:
\(-\frac{1}{2} \int du(xu) = -\frac{1}{2} \int x(400-100x^2)du\)

Now, we can integrate u:
\(-\frac{1}{2} \int x(400-100x^2)du = -\frac{1}{2} \int x(400-100x^2) du\)
\(-\frac{1}{2} \int x(400-100x^2) du = -\frac{1}{2} \int (400ux - 100ux^3) du\)
\(-\frac{1}{2} \int (400ux - 100ux^3) du = -\frac{1}{2} \left( 400 \int ux du - 100 \int ux^3 du \right)\)

Integrating \(ux\) with respect to u gives:
\(-\frac{1}{2} \left( 400 \int ux du - 100 \int ux^3 du \right) = -\frac{1}{2} \left( 400 \cdot \frac{u^2}{2} - 100 \cdot \frac{u^4}{4} \right)\)

Now, substitute the value of u back in terms of x:
\(-\frac{1}{2} \left( 400 \cdot \frac{u^2}{2} - 100 \cdot \frac{u^4}{4} \right) = -\frac{1}{2} \left( 400 \cdot \frac{(400-100x^2)^2}{2} - 100 \cdot \frac{(400-100x^2)^4}{4} \right)\)

Simplifying further:
\(-\frac{1}{2} \left( 400 \cdot \frac{(400-100x^2)^2}{2} - 100 \cdot \frac{(400-100x^2)^4}{4} \right) = -\frac{1}{2} \left( 200 \cdot (400-100x^2)^2 - 25 \cdot (400-100x^2)^4 \right) + C\)

So, the integral of \(x \sqrt{400-100x^2}\) dx is given by:
\(-\frac{1}{2} \left( 200 \cdot (400-100x^2)^2 - 25 \cdot (400-100x^2)^4 \right) + C\)