Solve ∫ (from - 2 to z) of 1/6(1+t^2)dt

$$\int _ { - 2 } ^ { z } \frac { 1 } { 6 } ( 1 + t ^ { 2 } ) d t$$

Evaluate

$$\frac{z^{3}+3z+14}{18}$$

Solution Steps

Use the distributive property to multiply \frac{1}{6} by 1+t^{2}.

$$\int _{-2}^{z}\frac{1}{6}+\frac{1}{6}t^{2}\mathrm{d}t$$

Evaluate the indefinite integral first.

$$\int \frac{1+t^{2}}{6}\mathrm{d}t$$

Integrate the sum term by term.

$$\int \frac{1}{6}\mathrm{d}t+\int \frac{t^{2}}{6}\mathrm{d}t$$

Factor out the constant in each of the terms.

$$\int \frac{1}{6}\mathrm{d}t+\frac{\int t^{2}\mathrm{d}t}{6}$$

Find the integral of \frac{1}{6} using the table of common integrals rule \int a\mathrm{d}t=at.

$$\frac{t+\int t^{2}\mathrm{d}t}{6}$$

Since \int t^{k}\mathrm{d}t=\frac{t^{k+1}}{k+1} for k\neq -1, replace \int t^{2}\mathrm{d}t with \frac{t^{3}}{3}. Multiply \frac{1}{6} times \frac{t^{3}}{3}.

$$\frac{t}{6}+\frac{t^{3}}{18}$$

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.

$$\frac{z}{6}+\frac{z^{3}}{18}-\left(\frac{1}{6}\left(-2\right)+\frac{\left(-2\right)^{3}}{18}\right)$$

Simplify.

$$\frac{z}{6}+\frac{z^{3}}{18}+\frac{7}{9}$$

Differentiate w.r.t. z

$$\frac{z^{2}+1}{6}$$