Solve ∫ _{-frac{sqrt{2}}{2}}^sqrt{2}(-m+8m^3-8m^5)dm

$$\int _ { - \frac { \sqrt { 2 } } { 2 } } ^ { \sqrt { 2 } } ( - m + 8 m ^ { 3 } - 8 m ^ { 5 } ) d m$$

Evaluate

$$-\frac{15}{4}=-3.75$$

Solution Steps

Evaluate the indefinite integral first.

$$\int -m+8m^{3}-8m^{5}\mathrm{d}m$$

Integrate the sum term by term.

$$\int -m\mathrm{d}m+\int 8m^{3}\mathrm{d}m+\int -8m^{5}\mathrm{d}m$$

Factor out the constant in each of the terms.

$$-\int m\mathrm{d}m+8\int m^{3}\mathrm{d}m-8\int m^{5}\mathrm{d}m$$

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

$$-\frac{m^{2}}{2}+8\int m^{3}\mathrm{d}m-8\int m^{5}\mathrm{d}m$$

Since \int m^{k}\mathrm{d}m=\frac{m^{k+1}}{k+1} for k\neq -1, replace \int m^{3}\mathrm{d}m with \frac{m^{4}}{4}. Multiply 8 times \frac{m^{4}}{4}.

$$-\frac{m^{2}}{2}+2m^{4}-8\int m^{5}\mathrm{d}m$$

Since \int m^{k}\mathrm{d}m=\frac{m^{k+1}}{k+1} for k\neq -1, replace \int m^{5}\mathrm{d}m with \frac{m^{6}}{6}. Multiply -8 times \frac{m^{6}}{6}.

$$-\frac{m^{2}}{2}+2m^{4}-\frac{4m^{6}}{3}$$

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.

$$-\frac{1}{2}\times \left(2^{\frac{1}{2}}\right)^{2}+2\times \left(2^{\frac{1}{2}}\right)^{4}-\frac{4}{3}\times \left(2^{\frac{1}{2}}\right)^{6}-\left(-\frac{1}{2}\left(-\frac{1}{2}\times 2^{\frac{1}{2}}\right)^{2}+2\left(-\frac{1}{2}\times 2^{\frac{1}{2}}\right)^{4}-\frac{4}{3}\left(-\frac{1}{2}\times 2^{\frac{1}{2}}\right)^{6}\right)$$

Simplify.

$$-\frac{15}{4}$$