The value of ∫_0^3 x√(x + 1) dx is

The correct answer is Option (4) → \(\displaystyle\frac{116}{15}\)

Let \(t = x + 1\). Then dt = dx and x = t − 1. With the change of limits:

    x = 0 gives t = 1; x = 3 gives t = 4

So

    \( \displaystyle \int_{0}^{3} x\sqrt{x + 1}\,dx = \int_{1}^{4} (t - 1)\,t^{1/2}\,dt = \int_{1}^{4} \left(t^{3/2} - t^{1/2}\right)dt \)

An antiderivative is

    \( \displaystyle G(t) = \frac{2}{5}t^{5/2} - \frac{2}{3}t^{3/2} \)

    \( \displaystyle G(4) = \frac{2}{5} \times 32 - \frac{2}{3} \times 8 = \frac{64}{5} - \frac{16}{3} = \frac{112}{15} \)

    \( \displaystyle G(1) = \frac{2}{5} - \frac{2}{3} = -\frac{4}{15} \)

    \( \displaystyle \int_{0}^{3} x\sqrt{x + 1}\,dx = \frac{112}{15} + \frac{4}{15} = \frac{116}{15} \)