Solve ∫ (from - 1 to 2) of 4x+16

$$\int _ { - 1 } ^ { 2 } 4 x + 16$$

Evaluate

$$54$$

Solution Steps

Evaluate the indefinite integral first.

$$\int 4x+16\mathrm{d}x$$

Integrate the sum term by term.

$$\int 4x\mathrm{d}x+\int 16\mathrm{d}x$$

Factor out the constant in each of the terms.

$$4\int x\mathrm{d}x+\int 16\mathrm{d}x$$

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 4 times \frac{x^{2}}{2}.

$$2x^{2}+\int 16\mathrm{d}x$$

Find the integral of 16 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$2x^{2}+16x$$

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.

$$2\times 2^{2}+16\times 2-\left(2\left(-1\right)^{2}+16\left(-1\right)\right)$$

Simplify.

$$54$$