Solve ∫ (from 0 to 5) of (x+2-x/2) wrt x

$$\int_{ 0 }^{ 5 } (x+2- \frac{ x }{ 2 } ) d x$$

Evaluate

$$\frac{65}{4}=16.25$$

Solution Steps

Combine x and -\frac{x}{2} to get \frac{1}{2}x.

$$\int _{0}^{5}\frac{1}{2}x+2\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int \frac{x}{2}+2\mathrm{d}x$$

Integrate the sum term by term.

$$\int \frac{x}{2}\mathrm{d}x+\int 2\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\frac{\int x\mathrm{d}x}{2}+\int 2\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x\mathrm{d}x with \frac{x^{2}}{2}. Multiply \frac{1}{2} times \frac{x^{2}}{2}.

$$\frac{x^{2}}{4}+\int 2\mathrm{d}x$$

Find the integral of 2 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{x^{2}}{4}+2x$$

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.

$$\frac{5^{2}}{4}+2\times 5-\left(\frac{0^{2}}{4}+2\times 0\right)$$

Simplify.

$$\frac{65}{4}$$