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

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

Evaluate

$$\frac{49}{3}\approx 16.333333333$$

Solution Steps

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

$$\int _{0}^{1}4x^{2}+12x+9\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int 4x^{2}+12x+9\mathrm{d}x$$

Integrate the sum term by term.

$$\int 4x^{2}\mathrm{d}x+\int 12x\mathrm{d}x+\int 9\mathrm{d}x$$

Factor out the constant in each of the terms.

$$4\int x^{2}\mathrm{d}x+12\int x\mathrm{d}x+\int 9\mathrm{d}x$$

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

$$\frac{4x^{3}}{3}+12\int x\mathrm{d}x+\int 9\mathrm{d}x$$

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

$$\frac{4x^{3}}{3}+6x^{2}+\int 9\mathrm{d}x$$

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

$$\frac{4x^{3}}{3}+6x^{2}+9x$$

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.

$$\frac{4}{3}\times 1^{3}+6\times 1^{2}+9\times 1-\left(\frac{4}{3}\times 0^{3}+6\times 0^{2}+9\times 0\right)$$

Simplify.

$$\frac{49}{3}$$