Solve ∫ (from - 1 to 3) of 2(x^3+4) wrt x

$$\int _ { - 1 } ^ { 3 } 2 ( x ^ { 3 } + 4 ) d x$$

Evaluate

$$72$$

Solution Steps

Use the distributive property to multiply 2 by x^{3}+4.

$$\int _{-1}^{3}2x^{3}+8\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int 2x^{3}+8\mathrm{d}x$$

Integrate the sum term by term.

$$\int 2x^{3}\mathrm{d}x+\int 8\mathrm{d}x$$

Factor out the constant in each of the terms.

$$2\int x^{3}\mathrm{d}x+\int 8\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{3}\mathrm{d}x with \frac{x^{4}}{4}. Multiply 2 times \frac{x^{4}}{4}.

$$\frac{x^{4}}{2}+\int 8\mathrm{d}x$$

Find the integral of 8 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{x^{4}}{2}+8x$$

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.

$$\frac{3^{4}}{2}+8\times 3-\left(\frac{\left(-1\right)^{4}}{2}+8\left(-1\right)\right)$$

Simplify.

$$72$$