Solve = ∫ (from 0 to 1) of z+t^2dt

$$= \int _ { 0 } ^ { 1 } z + t ^ { 2 } d t$$

Evaluate

$$z = \frac{1}{3} = 0.3333333333333333$$

Solution Steps

Evaluate the indefinite integral first.

$$\int z+t^{2}\mathrm{d}t$$

Integrate the sum term by term.

$$\int z\mathrm{d}t+\int t^{2}\mathrm{d}t$$

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

$$zt+\int t^{2}\mathrm{d}t$$

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}.

$$zt+\frac{t^{3}}{3}$$

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.

$$z\times 1+\frac{1^{3}}{3}-\left(z\times 0+\frac{0^{3}}{3}\right)$$

Simplify.

$$z+\frac{1}{3}$$

Differentiate w.r.t. z

$$1$$