Solve ∫ (from 1 to 2) of (4z-z^2-3) wrt z

$$\int _ { 1 } ^ { 2 } ( 4 z - z ^ { 2 } - 3 ) d z$$

Evaluate

$$\frac{2}{3}\approx 0.666666667$$

Solution Steps

Evaluate the indefinite integral first.

$$\int 4z-z^{2}-3\mathrm{d}z$$

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

Since \int z^{k}\mathrm{d}z=\frac{z^{k+1}}{k+1} for k\neq -1, replace \int z\mathrm{d}z with \frac{z^{2}}{2}. Multiply 4 times \frac{z^{2}}{2}.

$$2z^{2}-\int z^{2}\mathrm{d}z+\int -3\mathrm{d}z$$

Since \int z^{k}\mathrm{d}z=\frac{z^{k+1}}{k+1} for k\neq -1, replace \int z^{2}\mathrm{d}z with \frac{z^{3}}{3}. Multiply -1 times \frac{z^{3}}{3}.

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

Find the integral of -3 using the table of common integrals rule \int a\mathrm{d}z=az.

$$2z^{2}-\frac{z^{3}}{3}-3z$$

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.

$$2\times 2^{2}-\frac{2^{3}}{3}-3\times 2-\left(2\times 1^{2}-\frac{1^{3}}{3}-3\right)$$

Simplify.

$$\frac{2}{3}$$