Solve ∫ (from -2 to 2) of (x^4-3.75x^2-1) wrt x

$$\int_{ -2 }^{ 2 } ( { x }^{ 4 } -3.75 { x }^{ 2 } -1) d x$$

Evaluate

$$-11.2$$

Solution Steps

Evaluate the indefinite integral first.

$$\int x^{4}-\frac{15x^{2}}{4}-1\mathrm{d}x$$

Integrate the sum term by term.

$$\int x^{4}\mathrm{d}x+\int -\frac{15x^{2}}{4}\mathrm{d}x+\int -1\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int x^{4}\mathrm{d}x-\frac{15\int x^{2}\mathrm{d}x}{4}+\int -1\mathrm{d}x$$

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}.

$$\frac{x^{5}}{5}-\frac{15\int x^{2}\mathrm{d}x}{4}+\int -1\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{2}\mathrm{d}x with \frac{x^{3}}{3}. Multiply -3.75 times \frac{x^{3}}{3}.

$$\frac{x^{5}}{5}-\frac{5x^{3}}{4}+\int -1\mathrm{d}x$$

Find the integral of -1 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{x^{5}}{5}-\frac{5x^{3}}{4}-x$$

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.

$$\frac{2^{5}}{5}-\frac{5}{4}\times 2^{3}-2-\left(\frac{\left(-2\right)^{5}}{5}-\frac{5}{4}\left(-2\right)^{3}-\left(-2\right)\right)$$

Simplify.

$$-\frac{56}{5}$$