Question

Evaluate the definite integral. $$ \int_{0}^{1} \frac{x}{\sqrt{1+x}} d x $$

Short answer

The value of the integral is \(\frac{2\sqrt{2}}{3} - 2\sqrt{2} + \frac{2}{3} - 2 = \frac{2}{3} - \frac{4\sqrt{2}}{3}\).

Step-by-step solution

Substitution

Use the substitution \(u = 1 + x\). This implies that \(du = dx\) and when \(x = 0\), \(u = 1\) and when \(x = 1\), \(u = 2\). So, our new limits of integration are \(u = 1\) to \(u = 2\). This turns our integral into: \(\int_1^2\frac{u-1}{\sqrt{u}}du.\)

Simplification

This integral can now be split into two separate fractions: \(\int_1^2\frac{u}{\sqrt{u}}du - \int_1^2\frac{1}{\sqrt{u}}du\). This simplifies to: \(\int_1^2\sqrt{u}du - \int_1^2u^{-1/2}du.\)

Evaluation

These are now straightforward to integrate. The antiderivative of \(\sqrt{u}\) is \(\frac{2}{3}u^{3/2}\) and the antiderivative of \(u^{-1/2}\) is \(2u^{1/2}\). Evaluate these from 1 to 2 and subtract to get: \(\left[\frac{2}{3}(2)^{3/2}-2(2)^{1/2}\right] - \left[\frac{2}{3}(1)^{3/2}-2(1)^{1/2}\right]\). Simplifying these expressions gives the final answer.