Evaluate the integral (by interpreting it as the area between two curves) \int_{0}^{4} | \sqrt{x...

Question:

Evaluate the integral (by interpreting it as the area between two curves) {eq}\int_{0}^{4} | \sqrt{x + 2} - x| {/eq}

Area Between Functions With Integration:

When we have the definite integral set up with the absolute value function in the integrand, then basically it is calculating the area between the two curves. If there is just one curve, then the area is calculated with the x-axis.

Here the integral:

{eq}\int_{0}^{4} | \sqrt{x + 2} - x| dx {/eq}

is representing the area between the two functions

{eq}f(x)=\sqrt{x + 2} {/eq}

and

{eq}g(x)=x {/eq}

.

Now to evaluate this integral we need to get rid of the absolute sign, as follows:

In the interval:

{eq}0\le \:x\le \:2\\ \Rightarrow \left|\sqrt{x+2}-x\right|=\left(\sqrt{x+2}-x\right)\\ {/eq}

and in the interval:

{eq}2\le \:x\le \:4\\ \Rightarrow \left|\sqrt{x+2}-x\right|=\left(-\sqrt{x+2}+x\right)\\ {/eq}

So now we will split the given integral as follows:

{eq}\int _0^4\left|\sqrt{x+2}-x\right|dx=\int _0^2(\sqrt{x+2}-x) dx+\int _2^4(-\sqrt{x+2}+x) dx\\ =\int _0^2\sqrt{x+2}dx-\int _0^2xdx -\int _2^4\sqrt{x+2}dx+\int _2^4xdx\\ = \left[\frac{2}{3}(x+2)^{\frac{3}{2}}\right]^2_0 -\left[\frac{x^2}{2}\right]^2_0 -\left[\frac{2}{3}(x+2)^{\frac{3}{2}}\right]^4_2+\left[\frac{x^2}{2}\right]^4_2\\ =\frac{10}{3}-\frac{4\sqrt{2}}{3}-4\sqrt{6}+\frac{34}{3}\\ =2.983 {/eq}