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

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

Evaluate

$$0$$

Solution Steps

Evaluate the indefinite integral first.

$$\int x^{3}+\sin(x)\mathrm{d}x$$

Integrate the sum term by term.

$$\int x^{3}\mathrm{d}x+\int \sin(x)\mathrm{d}x$$

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}.

$$\frac{x^{4}}{4}+\int \sin(x)\mathrm{d}x$$

Use \int \sin(x)\mathrm{d}x=-\cos(x) from the table of common integrals to obtain the result.

$$\frac{x^{4}}{4}-\cos(x)$$

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.

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

Simplify.

$$0$$