Solve ∫ (from -2.5 to 2.5) of x^4-32 wrt x

$$\int_{ -2.5 }^{ 2.5 } { x }^{ 4 } -32 d x$$

Evaluate

$$-120.9375$$

Solution Steps

Evaluate the indefinite integral first.

$$\int x^{4}-32\mathrm{d}x$$

Integrate the sum term by term.

$$\int x^{4}\mathrm{d}x+\int -32\mathrm{d}x$$

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}.

$$\frac{x^{5}}{5}+\int -32\mathrm{d}x$$

Find the integral of -32 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{x^{5}}{5}-32x$$

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.

$$\frac{2.5^{5}}{5}-32\times 2.5-\left(\frac{\left(-2.5\right)^{5}}{5}-32\left(-2.5\right)\right)$$

Simplify.

$$-\frac{1935}{16}$$