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

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

Evaluate

$$\frac{206}{15}\approx 13.733333333$$

Solution Steps

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

$$\int _{0}^{2}\left(y^{2}\right)^{2}+2y^{2}+1\mathrm{d}y$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$\int _{0}^{2}y^{4}+2y^{2}+1\mathrm{d}y$$

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

Since \int y^{k}\mathrm{d}y=\frac{y^{k+1}}{k+1} for k\neq -1, replace \int y^{4}\mathrm{d}y with \frac{y^{5}}{5}.

$$\frac{y^{5}}{5}+2\int y^{2}\mathrm{d}y+\int 1\mathrm{d}y$$

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 2 times \frac{y^{3}}{3}.

$$\frac{y^{5}}{5}+\frac{2y^{3}}{3}+\int 1\mathrm{d}y$$

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

$$\frac{y^{5}}{5}+\frac{2y^{3}}{3}+y$$

Simplify.

$$y+\frac{2y^{3}}{3}+\frac{y^{5}}{5}$$

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.

$$2+\frac{2}{3}\times 2^{3}+\frac{2^{5}}{5}-\left(0+\frac{2}{3}\times 0^{3}+\frac{0^{5}}{5}\right)$$

Simplify.

$$\frac{206}{15}$$