https://file.aiquickdraw.com/mathbot/file/17272846581818vxx3p6d.jpg

The problem you uploaded asks to evaluate the definite integral:

\[
\int_{\frac{1}{2}}^{3} x^2 \sqrt{2x^3 - 4} \, dx
\]

To solve this, follow these steps:

### Step 1: Substitution
We'll start by making a substitution to simplify the square root expression. Let:

\[
u = 2x^3 - 4
\]

Now, differentiate \( u \) with respect to \( x \):

\[
\frac{du}{dx} = 6x^2
\]

Thus,

\[
du = 6x^2 \, dx
\]

We can now substitute into the integral. To change the limits of integration, we substitute the limits of \( x \) into the expression for \( u \):

- When \( x = \frac{1}{2} \):
\[
u = 2\left(\frac{1}{2}\right)^3 - 4 = 2\left(\frac{1}{8}\right) - 4 = \frac{1}{4} - 4 = -\frac{15}{4}
\]

- When \( x = 3 \):
\[
u = 2(3)^3 - 4 = 2(27) - 4 = 54 - 4 = 50
\]

Thus, the new limits for \( u \) are from \( u = -\frac{15}{4} \) to \( u = 50 \).

### Step 2: Rewrite the integral
We now rewrite the integral in terms of \( u \). Substituting \( u = 2x^3 - 4 \) and \( du = 6x^2 \, dx \), we get:

\[
\int_{-\frac{15}{4}}^{50} \frac{1}{6} \sqrt{u} \, du
\]

### Step 3: Evaluate the integral
The integral of \( \sqrt{u} \) is straightforward. We use the formula:

\[
\int \sqrt{u} \, du = \frac{2}{3} u^{3/2}
\]

Thus, the integral becomes:

\[
\frac{1}{6} \times \frac{2}{3} \left[ u^{3/2} \right]_{-\frac{15}{4}}^{50}
\]

Simplifying the constants:

\[
\frac{1}{6} \times \frac{2}{3} = \frac{1}{9}
\]

So, we have:

\[
\frac{1}{9} \left[ u^{3/2} \right]_{-\frac{15}{4}}^{50}
\]

### Step 4: Apply the limits
We now apply the limits to the expression \( u^{3/2} \):

- For \( u = 50 \):

\[
50^{3/2} = \left( \sqrt{50} \right)^3 = (7.071)^3 = 353.55
\]

- For \( u = -\frac{15}{4} \):

\[
\left( -\frac{15}{4} \right)^{3/2} \quad \text{(which will result in a complex number since the base is negative)}.
\]

However, evaluating integrals with complex numbers suggests the integral could have been defined differently. Would you like me to proceed with alternative methods, or check for other ways to handle the square root of negative values?

Evaluate the definite integral from 1/2 to 3 of x^2 sqrt(2x^3 - 4) dx.

Learn how to evaluate the definite integral of x^2 sqrt(2x^3 - 4) from 1/2 to 3 using the substitution method. This step-by-step solution covers integral techniques suitable for university-level calculus students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation