Solve ∫ (from - 36 to 36) of 3x^3-x

$$\int _ { - 36 } ^ { 36 } 3 x ^ { 3 } - x$$

Evaluate

$$0$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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 3 times \frac{x^{4}}{4}.

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

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

$$\frac{3x^{4}}{4}-\frac{x^{2}}{2}$$

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.

$$\frac{3}{4}\times 36^{4}-\frac{36^{2}}{2}-\left(\frac{3}{4}\left(-36\right)^{4}-\frac{\left(-36\right)^{2}}{2}\right)$$

Simplify.

$$0$$