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

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

Evaluate

$$12$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(1-x\right)^{2}.

$$\int _{0}^{3}x^{2}+1-2x+x^{2}\mathrm{d}x$$

Combine x^{2} and x^{2} to get 2x^{2}.

$$\int _{0}^{3}2x^{2}+1-2x\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int 2x^{2}+1-2x\mathrm{d}x$$

Integrate the sum term by term.

$$\int 2x^{2}\mathrm{d}x+\int 1\mathrm{d}x+\int -2x\mathrm{d}x$$

Factor out the constant in each of the terms.

$$2\int x^{2}\mathrm{d}x+\int 1\mathrm{d}x-2\int x\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{2}\mathrm{d}x with \frac{x^{3}}{3}. Multiply 2 times \frac{x^{3}}{3}.

$$\frac{2x^{3}}{3}+\int 1\mathrm{d}x-2\int x\mathrm{d}x$$

Find the integral of 1 using the table of common integrals rule \int a\mathrm{d}x=ax.

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

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

$$\frac{2x^{3}}{3}+x-x^{2}$$

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.

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

Simplify.

$$12$$