Solve ∫ (from 0 to 3) of (5x-3)^2 wrt x=

$$\int _ { 0 } ^ { 3 } ( 5 x - 3 ) ^ { 2 } d x =$$

Evaluate

$$117$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5x-3\right)^{2}.

$$\int _{0}^{3}25x^{2}-30x+9\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int 25x^{2}-30x+9\mathrm{d}x$$

Integrate the sum term by term.

$$\int 25x^{2}\mathrm{d}x+\int -30x\mathrm{d}x+\int 9\mathrm{d}x$$

Factor out the constant in each of the terms.

$$25\int x^{2}\mathrm{d}x-30\int x\mathrm{d}x+\int 9\mathrm{d}x$$

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

$$\frac{25x^{3}}{3}-30\int x\mathrm{d}x+\int 9\mathrm{d}x$$

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 -30 times \frac{x^{2}}{2}.

$$\frac{25x^{3}}{3}-15x^{2}+\int 9\mathrm{d}x$$

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

$$\frac{25x^{3}}{3}-15x^{2}+9x$$

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.

$$\frac{25}{3}\times 3^{3}-15\times 3^{2}+9\times 3-\left(\frac{25}{3}\times 0^{3}-15\times 0^{2}+9\times 0\right)$$

Simplify.

$$117$$