Solve ∫ (from 0 to 4) of (9-x^2)^2 wrt x

$$\int _ { 0 } ^ { 4 } ( 9 - x ^ { 2 } ) ^ { 2 } d x$$

Evaluate

$$\frac{724}{5}=144.8$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(9-x^{2}\right)^{2}.

$$\int _{0}^{4}81-18x^{2}+\left(x^{2}\right)^{2}\mathrm{d}x$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$\int _{0}^{4}81-18x^{2}+x^{4}\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int 81-18x^{2}+x^{4}\mathrm{d}x$$

Integrate the sum term by term.

$$\int 81\mathrm{d}x+\int -18x^{2}\mathrm{d}x+\int x^{4}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int 81\mathrm{d}x-18\int x^{2}\mathrm{d}x+\int x^{4}\mathrm{d}x$$

Find the integral of 81 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$81x-18\int x^{2}\mathrm{d}x+\int x^{4}\mathrm{d}x$$

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 -18 times \frac{x^{3}}{3}.

$$81x-6x^{3}+\int x^{4}\mathrm{d}x$$

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}.

$$81x-6x^{3}+\frac{x^{5}}{5}$$

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.

$$\frac{4^{5}}{5}-6\times 4^{3}+81\times 4-\left(\frac{0^{5}}{5}-6\times 0^{3}+81\times 0\right)$$

Simplify.

$$\frac{724}{5}$$