Solve ∫ (from 1 to 2) of (1+2y)^2 wrt y

$$\int _ { 1 } ^ { 2 } ( 1 + 2 y ) ^ { 2 } d y$$

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(1+2y\right)^{2}.

$$\int _{1}^{2}1+4y+4y^{2}\mathrm{d}y$$

Evaluate the indefinite integral first.

$$\int 1+4y+4y^{2}\mathrm{d}y$$

Integrate the sum term by term.

$$\int 1\mathrm{d}y+\int 4y\mathrm{d}y+\int 4y^{2}\mathrm{d}y$$

Factor out the constant in each of the terms.

$$\int 1\mathrm{d}y+4\int y\mathrm{d}y+4\int y^{2}\mathrm{d}y$$

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

$$y+4\int y\mathrm{d}y+4\int y^{2}\mathrm{d}y$$

Since \int y^{k}\mathrm{d}y=\frac{y^{k+1}}{k+1} for k\neq -1, replace \int y\mathrm{d}y with \frac{y^{2}}{2}. Multiply 4 times \frac{y^{2}}{2}.

$$y+2y^{2}+4\int y^{2}\mathrm{d}y$$

Since \int y^{k}\mathrm{d}y=\frac{y^{k+1}}{k+1} for k\neq -1, replace \int y^{2}\mathrm{d}y with \frac{y^{3}}{3}. Multiply 4 times \frac{y^{3}}{3}.

$$y+2y^{2}+\frac{4y^{3}}{3}$$

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.

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

Simplify.

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