Solve ∫ (from -1 to 5) of (1+3x) wrt x

$$\int_{ -1 }^{ 5 } (1+3x) d x$$

Evaluate

$$42$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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

$$x+3\int x\mathrm{d}x$$

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

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

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.

$$5+\frac{3}{2}\times 5^{2}-\left(-1+\frac{3}{2}\left(-1\right)^{2}\right)$$

Simplify.

$$42$$