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

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

Evaluate

$$-36$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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 -2 times \frac{x^{4}}{4}.

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

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}.

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

Find the integral of 4 using the table of common integrals rule \int a\mathrm{d}x=ax.

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

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.

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

Simplify.

$$-36$$