Solve ∫ (from 0 to 2) of (y+2)^2-y^4 wrt y

$$\int _ { 0 } ^ { 2 } ( y + 2 ) ^ { 2 } - y ^ { 4 } d y$$

Evaluate

$$\frac{184}{15}\approx 12.266666667$$

Solution Steps

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

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

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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}.

$$\frac{y^{3}}{3}+4\int y\mathrm{d}y+\int 4\mathrm{d}y-\int y^{4}\mathrm{d}y$$

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}.

$$\frac{y^{3}}{3}+2y^{2}+\int 4\mathrm{d}y-\int y^{4}\mathrm{d}y$$

Find the integral of 4 using the table of common integrals rule \int a\mathrm{d}y=ay.

$$\frac{y^{3}}{3}+2y^{2}+4y-\int y^{4}\mathrm{d}y$$

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}. Multiply -1 times \frac{y^{5}}{5}.

$$\frac{y^{3}}{3}+2y^{2}+4y-\frac{y^{5}}{5}$$

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.

$$\frac{2^{3}}{3}+2\times 2^{2}+4\times 2-\frac{2^{5}}{5}-\left(\frac{0^{3}}{3}+2\times 0^{2}+4\times 0-\frac{0^{5}}{5}\right)$$

Simplify.

$$\frac{184}{15}$$