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

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

Evaluate

$$\frac{56}{15}\approx 3.733333333$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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}.

$$\frac{y^{5}}{5}-\int y^{2}\mathrm{d}y-4\int y\mathrm{d}y+\int 4\mathrm{d}y$$

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}. Multiply -1 times \frac{y^{3}}{3}.

$$\frac{y^{5}}{5}-\frac{y^{3}}{3}-4\int y\mathrm{d}y+\int 4\mathrm{d}y$$

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}.

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

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

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

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.

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

Simplify.

$$\frac{56}{15}$$