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

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

Evaluate

$$\frac{3}{4}=0.75$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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}.

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

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

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

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.

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

Simplify.

$$\frac{3}{4}$$