Solve ∫ (from - 2 to 0) of (1/2x^4+1/4x^3-x) wrt x

$$\int _ { - 2 } ^ { 0 } ( \frac { 1 } { 2 } x ^ { 4 } + \frac { 1 } { 4 } x ^ { 3 } - x ) d x$$

Evaluate

$$\frac{21}{5}=4.2$$

Solution Steps

Evaluate the indefinite integral first.

$$\int \frac{x^{4}}{2}+\frac{x^{3}}{4}-x\mathrm{d}x$$

Integrate the sum term by term.

$$\int \frac{x^{4}}{2}\mathrm{d}x+\int \frac{x^{3}}{4}\mathrm{d}x+\int -x\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\frac{\int x^{4}\mathrm{d}x}{2}+\frac{\int x^{3}\mathrm{d}x}{4}-\int x\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{4}\mathrm{d}x with \frac{x^{5}}{5}. Multiply \frac{1}{2} times \frac{x^{5}}{5}.

$$\frac{x^{5}}{10}+\frac{\int x^{3}\mathrm{d}x}{4}-\int x\mathrm{d}x$$

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}. Multiply \frac{1}{4} times \frac{x^{4}}{4}.

$$\frac{x^{5}}{10}+\frac{x^{4}}{16}-\int x\mathrm{d}x$$

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 -1 times \frac{x^{2}}{2}.

$$\frac{x^{5}}{10}+\frac{x^{4}}{16}-\frac{x^{2}}{2}$$

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.

$$\frac{0^{5}}{10}+\frac{0^{4}}{16}-\frac{0^{2}}{2}-\left(\frac{\left(-2\right)^{5}}{10}+\frac{\left(-2\right)^{4}}{16}-\frac{\left(-2\right)^{2}}{2}\right)$$

Simplify.

$$\frac{21}{5}$$