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

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

Evaluate

$$\frac{14}{3}\approx 4.666666667$$

Solution Steps

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

$$\int _{0}^{2}x\left(\left(x^{2}\right)^{2}-6x^{2}+9\right)\mathrm{d}x$$

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

$$\int _{0}^{2}x\left(x^{4}-6x^{2}+9\right)\mathrm{d}x$$

Use the distributive property to multiply x by x^{4}-6x^{2}+9.

$$\int _{0}^{2}x^{5}-6x^{3}+9x\mathrm{d}x$$

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{5}\mathrm{d}x with \frac{x^{6}}{6}.

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

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{3}\mathrm{d}x with \frac{x^{4}}{4}. Multiply -6 times \frac{x^{4}}{4}.

$$\frac{x^{6}}{6}-\frac{3x^{4}}{2}+9\int x\mathrm{d}x$$

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 9 times \frac{x^{2}}{2}.

$$\frac{x^{6}}{6}-\frac{3x^{4}}{2}+\frac{9x^{2}}{2}$$

Simplify.

$$\frac{9x^{2}}{2}-\frac{3x^{4}}{2}+\frac{x^{6}}{6}$$

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.

$$\frac{9}{2}\times 2^{2}-\frac{3}{2}\times 2^{4}+\frac{2^{6}}{6}-\left(\frac{9}{2}\times 0^{2}-\frac{3}{2}\times 0^{4}+\frac{0^{6}}{6}\right)$$

Simplify.

$$\frac{14}{3}$$