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

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

Evaluate

$$\frac{80}{3}\approx 26.666666667$$

Solution Steps

Consider \left(3-x\right)\left(3+x\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

Calculate 3 to the power of 2 and get 9.

$$\int _{-1}^{3}9-x^{2}\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int 9-x^{2}\mathrm{d}x$$

Integrate the sum term by term.

$$\int 9\mathrm{d}x+\int -x^{2}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int 9\mathrm{d}x-\int x^{2}\mathrm{d}x$$

Find the integral of 9 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$9x-\int x^{2}\mathrm{d}x$$

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}.

$$9x-\frac{x^{3}}{3}$$

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.

$$9\times 3-\frac{3^{3}}{3}-\left(9\left(-1\right)-\frac{\left(-1\right)^{3}}{3}\right)$$

Simplify.

$$\frac{80}{3}$$