Solve ∫ (from t - 1 to 3) of (4z-8) wrt z

$$\int _ { t - 1 } ^ { 3 } ( 4 z - 8 ) d z$$

Evaluate

$$2\left(2-t\right)\left(t-4\right)$$

Solution Steps

Evaluate the indefinite integral first.

$$\int 4z-8\mathrm{d}z$$

Integrate the sum term by term.

$$\int 4z\mathrm{d}z+\int -8\mathrm{d}z$$

Factor out the constant in each of the terms.

$$4\int z\mathrm{d}z+\int -8\mathrm{d}z$$

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}.

$$2z^{2}+\int -8\mathrm{d}z$$

Find the integral of -8 using the table of common integrals rule \int a\mathrm{d}z=az.

$$2z^{2}-8z$$

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.

$$2\times 3^{2}-8\times 3-\left(2\left(t-1\right)^{2}-8\left(t-1\right)\right)$$

Simplify.

$$2\left(-4+t\right)\left(2-t\right)$$

Differentiate w.r.t. t

$$12-4t$$