Solve ∫ (from 0 to 1) of x^2(2-3x^2)^2 wrt x

$$\int _ { 0 } ^ { 1 } x ^ { 2 } ( 2 - 3 x ^ { 2 } ) ^ { 2 } d x$$

Evaluate

$$\frac{23}{105}\approx 0.219047619$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2-3x^{2}\right)^{2}.

$$\int _{0}^{1}x^{2}\left(4-12x^{2}+9\left(x^{2}\right)^{2}\right)\mathrm{d}x$$

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

$$\int _{0}^{1}x^{2}\left(4-12x^{2}+9x^{4}\right)\mathrm{d}x$$

Use the distributive property to multiply x^{2} by 4-12x^{2}+9x^{4}.

$$\int _{0}^{1}4x^{2}-12x^{4}+9x^{6}\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int 4x^{2}-12x^{4}+9x^{6}\mathrm{d}x$$

Integrate the sum term by term.

$$\int 4x^{2}\mathrm{d}x+\int -12x^{4}\mathrm{d}x+\int 9x^{6}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$4\int x^{2}\mathrm{d}x-12\int x^{4}\mathrm{d}x+9\int x^{6}\mathrm{d}x$$

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}.

$$\frac{4x^{3}}{3}-12\int x^{4}\mathrm{d}x+9\int x^{6}\mathrm{d}x$$

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}. Multiply -12 times \frac{x^{5}}{5}.

$$\frac{4x^{3}}{3}-\frac{12x^{5}}{5}+9\int x^{6}\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{6}\mathrm{d}x with \frac{x^{7}}{7}. Multiply 9 times \frac{x^{7}}{7}.

$$\frac{4x^{3}}{3}-\frac{12x^{5}}{5}+\frac{9x^{7}}{7}$$

Simplify.

$$\frac{9x^{7}}{7}-\frac{12x^{5}}{5}+\frac{4x^{3}}{3}$$

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.

$$\frac{9}{7}\times 1^{7}-\frac{12}{5}\times 1^{5}+\frac{4}{3}\times 1^{3}-\left(\frac{9}{7}\times 0^{7}-\frac{12}{5}\times 0^{5}+\frac{4}{3}\times 0^{3}\right)$$

Simplify.

$$\frac{23}{105}$$