Determine the volume of the region evaluating the double
integral:

Ask by Estrada Lee. in Kenya

Jan 24,2025

Question

Determine the volume of the region evaluating the double integral: $$ \int_{0}^{1} \int_{x^{2}}^{x}(x y) d y d x $$

UpStudy AI · Accepted Answer

To determine the volume represented by the double integral:

$$
\int_{0}^{1} \int_{x^{2}}^{x} x y \, dy \, dx,
$$

we'll evaluate the integral step by step.

### Step 1: Evaluate the Inner Integral

First, hold $$ x $$ constant and integrate with respect to $$ y $$:

$$
\int_{x^{2}}^{x} x y \, dy
$$

Since $$ x $$ is constant with respect to $$ y $$, we can factor it out:

$$
x \int_{x^{2}}^{x} y \, dy
$$

Now, integrate $$ y $$ with respect to $$ y $$:

$$
x \left[ \frac{1}{2} y^2 \right]_{x^{2}}^{x} = x \left( \frac{1}{2} x^2 - \frac{1}{2} (x^2)^2 \right) = x \left( \frac{1}{2} x^2 - \frac{1}{2} x^4 \right) = \frac{1}{2} x^3 - \frac{1}{2} x^5
$$

### Step 2: Evaluate the Outer Integral

Now, integrate the result with respect to $$ x $$:

$$
\int_{0}^{1} \left( \frac{1}{2} x^3 - \frac{1}{2} x^5 \right) dx = \frac{1}{2} \int_{0}^{1} x^3 \, dx - \frac{1}{2} \int_{0}^{1} x^5 \, dx
$$

Calculate each integral separately:

$$
\frac{1}{2} \left[ \frac{1}{4} x^4 \right]_{0}^{1} - \frac{1}{2} \left[ \frac{1}{6} x^6 \right]_{0}^{1} = \frac{1}{2} \left( \frac{1}{4} - 0 \right) - \frac{1}{2} \left( \frac{1}{6} - 0 \right)
$$

Simplify:

$$
\frac{1}{2} \cdot \frac{1}{4} - \frac{1}{2} \cdot \frac{1}{6} = \frac{1}{8} - \frac{1}{12} = \frac{3}{24} - \frac{2}{24} = \frac{1}{24}
$$

### Conclusion

The volume of the region is:

$$
\boxed{\dfrac{1}{24}}
$$
The volume of the region is $$ \frac{1}{24} $$.

Determine the volume of the region evaluating the double
integral:

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