Solve ∫ (from 0 to 4) of (3-4x)^2 wrt x

$$\int _ { 0 } ^ { 4 } ( 3 - 4 x ) ^ { 2 } d x$$

Evaluate

$$\frac{556}{3}\approx 185.333333333$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3-4x\right)^{2}.

$$\int _{0}^{4}9-24x+16x^{2}\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int 9-24x+16x^{2}\mathrm{d}x$$

Integrate the sum term by term.

$$\int 9\mathrm{d}x+\int -24x\mathrm{d}x+\int 16x^{2}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int 9\mathrm{d}x-24\int x\mathrm{d}x+16\int x^{2}\mathrm{d}x$$

Find the integral of 9 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$9x-24\int x\mathrm{d}x+16\int x^{2}\mathrm{d}x$$

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 -24 times \frac{x^{2}}{2}.

$$9x-12x^{2}+16\int x^{2}\mathrm{d}x$$

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}.

$$9x-12x^{2}+\frac{16x^{3}}{3}$$

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.

$$9\times 4-12\times 4^{2}+\frac{16}{3}\times 4^{3}-\left(9\times 0-12\times 0^{2}+\frac{16}{3}\times 0^{3}\right)$$

Simplify.

$$\frac{556}{3}$$