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

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

Evaluate

$$\frac{38}{15}\approx 2.533333333$$

Solution Steps

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

$$\int _{2}^{3}\left(x^{2}\right)^{2}-4x^{2}x+4x^{2}\mathrm{d}x$$

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

$$\int _{2}^{3}x^{4}-4x^{2}x+4x^{2}\mathrm{d}x$$

To multiply powers of the same base, add their exponents. Add 2 and 1 to get 3.

$$\int _{2}^{3}x^{4}-4x^{3}+4x^{2}\mathrm{d}x$$

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{4}\mathrm{d}x with \frac{x^{5}}{5}.

$$\frac{x^{5}}{5}-4\int x^{3}\mathrm{d}x+4\int x^{2}\mathrm{d}x$$

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

$$\frac{x^{5}}{5}-x^{4}+4\int x^{2}\mathrm{d}x$$

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

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

Simplify.

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

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.

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

Simplify.

$$\frac{38}{15}$$