To solve the integral [tex]\(\int_a^{\sqrt{7}} t \left( t^2 + 1 \right)^{\frac{1}{3}} \, dt\)[/tex], we follow these steps:
1. Identify the Integrand: Here, we need to integrate the function [tex]\( t \left( t^2 + 1 \right)^{\frac{1}{3}} \)[/tex].
2. Substitution: We use a substitution to simplify the integral. Let [tex]\( u = t^2 + 1 \)[/tex]. Then, differentiate both sides with respect to [tex]\( t \)[/tex]:
[tex]\[
\frac{du}{dt} = 2t \quad \Rightarrow \quad du = 2t \, dt \quad \Rightarrow \quad t \, dt = \frac{1}{2} \, du
\][/tex]
3. Transform the Integral: Substituting [tex]\( u = t^2 + 1 \)[/tex] and [tex]\( t \, dt = \frac{1}{2} \, du \)[/tex] into the integral:
[tex]\[
\int_a^{\sqrt{7}} t \left( t^2 + 1 \right)^{\frac{1}{3}} \, dt = \int_{u(a)}^{u(\sqrt{7})} \frac{1}{2} u^{\frac{1}{3}} \, du
\][/tex]
where [tex]\( u(a) = a^2 + 1 \)[/tex] and [tex]\( u(\sqrt{7}) = (\sqrt{7})^2 + 1 = 7 + 1 = 8 \)[/tex].
4. Integrate: Now the integral is in terms of [tex]\( u \)[/tex]:
[tex]\[
\frac{1}{2} \int_{a^2 + 1}^{8} u^{\frac{1}{3}} \, du
\][/tex]
To integrate [tex]\( u^{\frac{1}{3}} \)[/tex], use the power rule for integration:
[tex]\[
\int u^{\frac{1}{3}} \, du = \frac{u^{\frac{1}{3} + 1}}{\frac{1}{3} + 1} = \frac{u^{\frac{4}{3}}}{\frac{4}{3}} = \frac{3}{4} u^{\frac{4}{3}}
\][/tex]
5. Apply the Limits: Include the limits [tex]\( a^2 + 1 \)[/tex] to [tex]\( 8 \)[/tex]:
[tex]\[
\frac{1}{2} \left[ \frac{3}{4} u^{\frac{4}{3}} \right]_{a^2 + 1}^{8} = \frac{3}{8} \left[ u^{\frac{4}{3}} \right]_{a^2 + 1}^{8}
\][/tex]
Evaluating this at the bounds:
[tex]\[
\frac{3}{8} \left( 8^{\frac{4}{3}} - (a^2 + 1)^{\frac{4}{3}} \right)
\][/tex]
6. Compute the Powers: Calculate the values inside the brackets:
[tex]\[
8^{\frac{4}{3}} = \left( 2^3 \right)^{\frac{4}{3}} = 2^4 = 16
\][/tex]
Thus, we have:
[tex]\[
\frac{3}{8} \left( 16 - (a^2 + 1)^{\frac{4}{3}} \right)
\][/tex]
7. Simplify: Distribute:
[tex]\[
\frac{3}{8} \cdot 16 - \frac{3}{8} \cdot (a^2 + 1)^{\frac{4}{3}} = 6 - \frac{3}{8} (a^2 + 1)^{\frac{4}{3}}
\][/tex]
8. Final Answer: The result of the integral is:
[tex]\[
6 - 0.375 (a^2 + 1)^{\frac{4}{3}}
\][/tex]
