Solve ∫ (from 0 to 1) of (1-4x)^2 wrt x

$$\int _ { 0 } ^ { 1 } ( 1 - 4 x ) ^ { 2 } d x$$

Evaluate

$$\frac{7}{3}\approx 2.333333333$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(1-4x\right)^{2}.

$$\int _{0}^{1}1-8x+16x^{2}\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int 1-8x+16x^{2}\mathrm{d}x$$

Integrate the sum term by term.

$$\int 1\mathrm{d}x+\int -8x\mathrm{d}x+\int 16x^{2}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int 1\mathrm{d}x-8\int x\mathrm{d}x+16\int x^{2}\mathrm{d}x$$

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

$$x-8\int x\mathrm{d}x+16\int x^{2}\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x\mathrm{d}x with \frac{x^{2}}{2}. Multiply -8 times \frac{x^{2}}{2}.

$$x-4x^{2}+16\int x^{2}\mathrm{d}x$$

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

$$x-4x^{2}+\frac{16x^{3}}{3}$$

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.

$$1-4\times 1^{2}+\frac{16}{3}\times 1^{3}-\left(0-4\times 0^{2}+\frac{16}{3}\times 0^{3}\right)$$

Simplify.

$$\frac{7}{3}$$