Solve ∫ (from - 12 to 4) of (6-y-1/8y^2) wrt y

$$\int _ { - 12 } ^ { 4 } ( 6 - y - \frac { 1 } { 8 } y ^ { 2 } ) d y$$

Evaluate

$$\frac{256}{3}\approx 85.333333333$$

Solution Steps

Evaluate the indefinite integral first.

$$\int 6-y-\frac{y^{2}}{8}\mathrm{d}y$$

Integrate the sum term by term.

$$\int 6\mathrm{d}y+\int -y\mathrm{d}y+\int -\frac{y^{2}}{8}\mathrm{d}y$$

Factor out the constant in each of the terms.

$$\int 6\mathrm{d}y-\int y\mathrm{d}y-\frac{\int y^{2}\mathrm{d}y}{8}$$

Find the integral of 6 using the table of common integrals rule \int a\mathrm{d}y=ay.

$$6y-\int y\mathrm{d}y-\frac{\int y^{2}\mathrm{d}y}{8}$$

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 -1 times \frac{y^{2}}{2}.

$$6y-\frac{y^{2}}{2}-\frac{\int y^{2}\mathrm{d}y}{8}$$

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 -\frac{1}{8} times \frac{y^{3}}{3}.

$$6y-\frac{y^{2}}{2}-\frac{y^{3}}{24}$$

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.

$$6\times 4-\frac{4^{2}}{2}-\frac{4^{3}}{24}-\left(6\left(-12\right)-\frac{\left(-12\right)^{2}}{2}-\frac{\left(-12\right)^{3}}{24}\right)$$

Simplify.

$$\frac{256}{3}$$