# Evaluate the integral. {eq}\displaystyle \int_{1}^{16} \frac{x - 5}{\sqrt x}\ dx {/eq}.

## Question:

Evaluate the integral.

{eq}\displaystyle \int_{1}^{16} \frac{x - 5}{\sqrt x}\ dx {/eq}.

## Definite Integral in Calculus:

The definite integral is used to compute an exact area. If a function {eq}f(x) {/eq} is continuous on the interval {eq}\left[ a,b \right] {/eq} then the definite integral of {eq}f(x) {/eq} from {eq}a {/eq} to {eq}b {/eq} is {eq}\displaystyle \int_{a}^{b} f(x) \ dx {/eq}.

To solve this problem, we'll use the integral sum rule. Which states that:{eq}\displaystyle \; \int \left( f(x) \pm g(x) \right) \ dx = \int f(x) \ dx \pm \int g(x) \ dx {/eq}

Next, we'll compute the boundaries.

We are given:

{eq}\displaystyle \int_{1}^{16} \frac{x - 5}{\sqrt x}\ dx {/eq}

Compute the indefinite integral:

{eq}=\displaystyle \int \frac{x - 5}{\sqrt{x}} \, \mathrm{d}x {/eq}

{eq}= \displaystyle \int \left( \frac{x }{\sqrt{x}} -\frac{ 5}{\sqrt{x}} \right) \, \mathrm{d}x {/eq}

{eq}= \displaystyle \int \left( \sqrt{x} -\frac{ 5}{\sqrt{x}} \right) \, \mathrm{d}x {/eq}

{eq}= \displaystyle \int \left( x^{1/2} -5x^{-1/2} \right) \, \mathrm{d}x {/eq}

Apply the integral sum rule:

{eq}= \displaystyle \int x^{1/2} \ dx - \int 5x^{-1/2} \, \mathrm{d}x {/eq}

Take the constant out:

{eq}= \displaystyle \int x^{1/2} \ dx - 5 \int x^{-1/2} \, \mathrm{d}x {/eq}

{eq}= \displaystyle \dfrac{ x^{1/2+1} }{\dfrac{1}{2}+1 } - 5 \dfrac{ x^{-1/2+1} }{-\dfrac{1}{2}+1 } {/eq}

{eq}= \displaystyle \dfrac{2}{3} x^{3/2} - 10 x^{1/2} {/eq}

{eq}= \displaystyle \dfrac{2}{3} x^{3/2} - 10 x^{1/2} + C {/eq}

Compute the boundaries:

{eq}= \displaystyle \dfrac{8}{3}+ \dfrac{28}{3} {/eq}

{eq}= \displaystyle 12 {/eq}

Therefore the solution is: {eq}\displaystyle \int_{1}^{16} \frac{x - 5}{\sqrt{x}} \, \mathrm{d}x = 12 {/eq}