The integral given in the image is:

$\int_{-\frac{1}{2}}^{\frac{1}{2}} (3x+1)^7 \, dx$

To solve this, let's use substitution. Set:

$u = 3x + 1 \quad \Rightarrow \quad du = 3dx \quad \Rightarrow \quad dx = \frac{du}{3}$

Also, the limits of integration change with the substitution:

• When $x = -\frac{1}{2}$, $u = 3\left(-\frac{1}{2}\right) + 1 = \frac{1}{2}$
• When $x = \frac{1}{2}$, $u = 3\left(\frac{1}{2}\right) + 1 = \frac{5}{2}$

Thus, the integral becomes:

$\int_{\frac{1}{2}}^{\frac{5}{2}} u^7 \cdot \frac{du}{3}$

Factor out $\frac{1}{3}$:

$\frac{1}{3} \int_{\frac{1}{2}}^{\frac{5}{2}} u^7 \, du$

Now integrate $u^7$:

$\frac{1}{3} \left[ \frac{u^8}{8} \right]_{\frac{1}{2}}^{\frac{5}{2}}$

Now, substitute the limits of integration:

$\frac{1}{3} \left( \frac{\left( \frac{5}{2} \right)^8}{8} - \frac{\left( \frac{1}{2} \right)^8}{8} \right)$

Simplify both terms:

• $\left( \frac{5}{2} \right)^8 = \frac{390625}{256}$
• $\left( \frac{1}{2} \right)^8 = \frac{1}{256}$

Substitute these values back:

$\frac{1}{3} \left( \frac{390625}{2048} - \frac{1}{2048} \right) = \frac{1}{3} \cdot \frac{390624}{2048} = \frac{390624}{6144}$

Thus, the final value of the integral is:

$\frac{390624}{6144}$

Would you like further details on any of the steps?

Here are 5 related questions you can explore:

1. How does substitution work in more complex integrals?
2. How can you approach integration for higher powers?
3. What are other common methods to solve definite integrals?
4. How does the Fundamental Theorem of Calculus apply to definite integrals?
5. What are the techniques for approximating definite integrals?

Tip: Always remember to change the limits of integration when performing a substitution in definite integrals.