Solve ∫ (from 0 to sqrt { 2) of }(2-x^2)^2 wrt x

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

Evaluate

$$\frac{32\sqrt{2}}{15}\approx 3.016988933$$

Solution Steps

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

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

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

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

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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

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

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 -4 times \frac{x^{3}}{3}.

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

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}.

$$4x-\frac{4x^{3}}{3}+\frac{x^{5}}{5}$$

Simplify.

$$\frac{x^{5}}{5}-\frac{4x^{3}}{3}+4x$$

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.

$$\frac{1}{5}\times \left(2^{\frac{1}{2}}\right)^{5}-\frac{4}{3}\times \left(2^{\frac{1}{2}}\right)^{3}+4\times 2^{\frac{1}{2}}-\left(\frac{0^{5}}{5}-\frac{4}{3}\times 0^{3}+4\times 0\right)$$

Simplify.

$$\frac{32\sqrt{2}}{15}$$