Solve ∫ (from 0 to 16) of sqrt{x}+sqrt{y-4}

$$\int _ { 0 } ^ { 16 } \sqrt { x } + \sqrt { y - 4 }$$

Evaluate

$$16\sqrt{y-4}+\frac{128}{3}$$

Solution Steps

Evaluate the indefinite integral first.

$$\int \sqrt{x}+\sqrt{y-4}\mathrm{d}x$$

Integrate the sum term by term.

$$\int \sqrt{x}\mathrm{d}x+\int \sqrt{y-4}\mathrm{d}x$$

Rewrite \sqrt{x} as x^{\frac{1}{2}}. Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{\frac{1}{2}}\mathrm{d}x with \frac{x^{\frac{3}{2}}}{\frac{3}{2}}. Simplify.

$$\frac{2x^{\frac{3}{2}}}{3}+\int \sqrt{y-4}\mathrm{d}x$$

Find the integral of \sqrt{y-4} using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{2x^{\frac{3}{2}}}{3}+\sqrt{y-4}x$$

The definite integral is the antiderivative of the expression evaluated at the upper limit of integration minus the antiderivative evaluated at the lower limit of integration.

$$\frac{2}{3}\times 16^{\frac{3}{2}}+\left(y-4\right)^{\frac{1}{2}}\times 16-\left(\frac{2}{3}\times 0^{\frac{3}{2}}+\left(y-4\right)^{\frac{1}{2}}\times 0\right)$$

Simplify.

$$\frac{128}{3}+16\sqrt{y-4}$$

Differentiate w.r.t. y

$$\frac{8}{\sqrt{y-4}}$$