Solve ∫ (from 0 to 4) of (4-(y-1)^2)(4-y) wrt y

$$\int _ { 0 } ^ { 4 } ( 4 - ( y - 1 ) ^ { 2 } ) ( 4 - y ) d y$$

Evaluate

$$24$$

Solution Steps

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

$$\int _{0}^{4}\left(4-\left(y^{2}-2y+1\right)\right)\left(4-y\right)\mathrm{d}y$$

To find the opposite of y^{2}-2y+1, find the opposite of each term.

$$\int _{0}^{4}\left(4-y^{2}+2y-1\right)\left(4-y\right)\mathrm{d}y$$

Subtract 1 from 4 to get 3.

$$\int _{0}^{4}\left(3-y^{2}+2y\right)\left(4-y\right)\mathrm{d}y$$

Use the distributive property to multiply 3-y^{2}+2y by 4-y and combine like terms.

$$\int _{0}^{4}12+5y-6y^{2}+y^{3}\mathrm{d}y$$

Evaluate the indefinite integral first.

$$\int 12+5y-6y^{2}+y^{3}\mathrm{d}y$$

Integrate the sum term by term.

$$\int 12\mathrm{d}y+\int 5y\mathrm{d}y+\int -6y^{2}\mathrm{d}y+\int y^{3}\mathrm{d}y$$

Factor out the constant in each of the terms.

$$\int 12\mathrm{d}y+5\int y\mathrm{d}y-6\int y^{2}\mathrm{d}y+\int y^{3}\mathrm{d}y$$

Find the integral of 12 using the table of common integrals rule \int a\mathrm{d}y=ay.

$$12y+5\int y\mathrm{d}y-6\int y^{2}\mathrm{d}y+\int y^{3}\mathrm{d}y$$

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

$$12y+\frac{5y^{2}}{2}-6\int y^{2}\mathrm{d}y+\int y^{3}\mathrm{d}y$$

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 -6 times \frac{y^{3}}{3}.

$$12y+\frac{5y^{2}}{2}-2y^{3}+\int y^{3}\mathrm{d}y$$

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

$$12y+\frac{5y^{2}}{2}-2y^{3}+\frac{y^{4}}{4}$$

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.

$$12\times 4+\frac{5}{2}\times 4^{2}-2\times 4^{3}+\frac{4^{4}}{4}-\left(12\times 0+\frac{5}{2}\times 0^{2}-2\times 0^{3}+\frac{0^{4}}{4}\right)$$

Simplify.

$$24$$