Solve ∫ 2e^-0*1t}dt

$$\int{ 2 { e }^{ -0 \cdot 1t } }d t$$

Evaluate

$$2t+С$$

Solution Steps

Multiply 0 and 1 to get 0.

$$\int 2e^{0t}\mathrm{d}t$$

Anything times zero gives zero.

$$\int 2e^{0}\mathrm{d}t$$

Calculate e to the power of 0 and get 1.

$$\int 2\times 1\mathrm{d}t$$

Multiply 2 and 1 to get 2.

$$\int 2\mathrm{d}t$$

Find the integral of 2 using the table of common integrals rule \int a\mathrm{d}t=at.

$$2t$$

If F\left(t\right) is an antiderivative of f\left(t\right), then the set of all antiderivatives of f\left(t\right) is given by F\left(t\right)+C. Therefore, add the constant of integration C\in \mathrm{R} to the result.

$$2t+С$$

Differentiate w.r.t. t

$$2$$