Solve ∫ (from - 5 to 5) of x^2[x+1/2]

$$\int _ { - 5 } ^ { 5 } x ^ { 2 } [ x + \frac { 1 } { 2 } ]$$

Evaluate

$$\frac{125}{3}\approx 41.666666667$$

Solution Steps

Use the distributive property to multiply x^{2} by x+\frac{1}{2}.

$$\int _{-5}^{5}x^{3}+\frac{1}{2}x^{2}\mathrm{d}x$$

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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}.

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

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{2}\mathrm{d}x with \frac{x^{3}}{3}. Multiply \frac{1}{2} times \frac{x^{3}}{3}.

$$\frac{x^{4}}{4}+\frac{x^{3}}{6}$$

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.

$$\frac{5^{4}}{4}+\frac{5^{3}}{6}-\left(\frac{\left(-5\right)^{4}}{4}+\frac{\left(-5\right)^{3}}{6}\right)$$

Simplify.

$$\frac{125}{3}$$