Question

In Exercises \(53-66,\) make a \(u\) -substitution and integrate from \(u(a)\) to \(u(b) .\) $$\int_{0}^{1} \sqrt{t^{5}+2 t}\left(5 t^{4}+2\right) d t$$

Short answer

The integral evaluates to \(2\sqrt{3}\).

Step-by-step solution

Identify a suitable substitution

We see powers and roots of functions inside the integral, which is often a sign that a \(u\)-substitution could be useful. In this case, let \(u = t^{5}+2 t\). This choice makes the integrand simplify considerably after substitution.

Differentiate u with respect to t

Differentiating \(u\) with respect to \(t\) gives us \(du/dt = 5t^{4} + 2\). We need this to substitute for \(dt\) in the integral. Rearranging this gives \(dt = du/(5t^{4} + 2)\)

Substitute u and dt in the integral

Now we can substitute \(u\) for \(t^{5}+2t\) and \(dt\) for \(du/(5t^{4} + 2)\) in the integral. Doing so results in \(\int \sqrt{u} du\).

Adjust the limits of integration

We initially integrated from \(t=0\) to \(t=1\). We must express these limits in terms of \(u\). When \(t = 0\), \(u = 0^{5} + 2*0 = 0\). And when \(t = 1\), \(u = 1^{5} + 2*1 = 3\). Thus, the integral becomes \(\int_{0}^{3} \sqrt{u} du\).

Evaluate the integral

The integral of \(\sqrt{u}\) with respect to \(u\) from 0 to 3 is \(2/3*[u^{3/2}]_{0}^{3}\), which evaluates to \(2/3*[3\sqrt{3} - 0] = 2\sqrt{3}\).