Solve ∫ _{e^x}^x(3t+1)dt

$$\int _ { e ^ { x } } ^ { x } ( 3 t + 1 ) d t$$

Evaluate

$$-\frac{3e^{2x}}{2}-e^{x}+\frac{3x^{2}}{2}+x$$

Solution Steps

Evaluate the indefinite integral first.

$$\int 3t+1\mathrm{d}t$$

Integrate the sum term by term.

$$\int 3t\mathrm{d}t+\int 1\mathrm{d}t$$

Factor out the constant in each of the terms.

$$3\int t\mathrm{d}t+\int 1\mathrm{d}t$$

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

$$\frac{3t^{2}}{2}+\int 1\mathrm{d}t$$

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

$$\frac{3t^{2}}{2}+t$$

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.

$$\frac{3}{2}x^{2}+x-\left(\frac{3}{2}\left(e^{x}\right)^{2}+e^{x}\right)$$

Simplify.

$$\frac{3x^{2}}{2}+x-\frac{3e^{2x}}{2}-e^{x}$$

Differentiate w.r.t. x

$$3x-3e^{2x}-e^{x}+1$$