Solve ∫ (from - 4 to 7) of (x^2-1)(x+1) wrt x

$$\int _ { - 4 } ^ { 7 } ( x ^ { 2 } - 1 ) ( x + 1 ) d x$$

Evaluate

$$\frac{7733}{12}\approx 644.416666667$$

Solution Steps

Use the distributive property to multiply x^{2}-1 by x+1.

$$\int _{-4}^{7}x^{3}+x^{2}-x-1\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int x^{3}+x^{2}-x-1\mathrm{d}x$$

Integrate the sum term by term.

$$\int x^{3}\mathrm{d}x+\int x^{2}\mathrm{d}x+\int -x\mathrm{d}x+\int -1\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int x^{3}\mathrm{d}x+\int x^{2}\mathrm{d}x-\int x\mathrm{d}x+\int -1\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{3}\mathrm{d}x with \frac{x^{4}}{4}.

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

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}.

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

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

$$\frac{x^{4}}{4}+\frac{x^{3}}{3}-\frac{x^{2}}{2}+\int -1\mathrm{d}x$$

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

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

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.

$$\frac{7^{4}}{4}+\frac{7^{3}}{3}-\frac{7^{2}}{2}-7-\left(\frac{\left(-4\right)^{4}}{4}+\frac{\left(-4\right)^{3}}{3}-\frac{\left(-4\right)^{2}}{2}-\left(-4\right)\right)$$

Simplify.

$$\frac{7733}{12}$$