Solve ∫ (from - 4 to 1) of (4-x^2-3x) wrt x

$$\int _ { - 4 } ^ { 1 } ( 4 - x ^ { 2 } - 3 x ) d x$$

Evaluate

$$\frac{125}{6}\approx 20.833333333$$

Solution Steps

Evaluate the indefinite integral first.

$$\int 4-x^{2}-3x\mathrm{d}x$$

Integrate the sum term by term.

$$\int 4\mathrm{d}x+\int -x^{2}\mathrm{d}x+\int -3x\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int 4\mathrm{d}x-\int x^{2}\mathrm{d}x-3\int x\mathrm{d}x$$

Find the integral of 4 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$4x-\int x^{2}\mathrm{d}x-3\int x\mathrm{d}x$$

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

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

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 -3 times \frac{x^{2}}{2}.

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

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.

$$4\times 1-\frac{1^{3}}{3}-\frac{3}{2}\times 1^{2}-\left(4\left(-4\right)-\frac{\left(-4\right)^{3}}{3}-\frac{3}{2}\left(-4\right)^{2}\right)$$

Simplify.

$$\frac{125}{6}$$