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

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

Evaluate

$$-\frac{74}{3}\approx -24.666666667$$

Solution Steps

To find the opposite of 3+x, find the opposite of each term.

$$\int _{-1}^{1}\left(-3-x\right)\left(x+4\right)\mathrm{d}x$$

Apply the distributive property by multiplying each term of -3-x by each term of x+4.

$$\int _{-1}^{1}-3x-12-x^{2}-4x\mathrm{d}x$$

Combine -3x and -4x to get -7x.

$$\int _{-1}^{1}-7x-12-x^{2}\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int -7x-12-x^{2}\mathrm{d}x$$

Integrate the sum term by term.

$$\int -7x\mathrm{d}x+\int -12\mathrm{d}x+\int -x^{2}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$-7\int x\mathrm{d}x+\int -12\mathrm{d}x-\int x^{2}\mathrm{d}x$$

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 -7 times \frac{x^{2}}{2}.

$$-\frac{7x^{2}}{2}+\int -12\mathrm{d}x-\int x^{2}\mathrm{d}x$$

Find the integral of -12 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$-\frac{7x^{2}}{2}-12x-\int x^{2}\mathrm{d}x$$

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}.

$$-\frac{7x^{2}}{2}-12x-\frac{x^{3}}{3}$$

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.

$$-\frac{7}{2}\times 1^{2}-12-\frac{1^{3}}{3}-\left(-\frac{7}{2}\left(-1\right)^{2}-12\left(-1\right)-\frac{\left(-1\right)^{3}}{3}\right)$$

Simplify.

$$-\frac{74}{3}$$