Solve ∫ (from 0 to frac { 1) of {2}}(x^3+2) wrt x

$$\int _ { 0 } ^ { \frac { 1 } { 2 } } ( x ^ { 3 } + 2 ) d x$$

Evaluate

$$\frac{65}{64}=1.015625$$

Solution Steps

Evaluate the indefinite integral first.

$$\int x^{3}+2\mathrm{d}x$$

Integrate the sum term by term.

$$\int x^{3}\mathrm{d}x+\int 2\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{3}\mathrm{d}x with \frac{x^{4}}{4}.

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

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

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

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.

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

Simplify.

$$\frac{65}{64}$$