Solve ∫ (from - 2 to 6) of (2x^3-x^2+4)

$$\int _ { - 2 } ^ { 6 } ( 2 x ^ { 3 } - x ^ { 2 } + 4 )$$

Evaluate

$$\frac{1792}{3}\approx 597.333333333$$

Solution Steps

Evaluate the indefinite integral first.

$$\int 2x^{3}-x^{2}+4\mathrm{d}x$$

Integrate the sum term by term.

$$\int 2x^{3}\mathrm{d}x+\int -x^{2}\mathrm{d}x+\int 4\mathrm{d}x$$

Factor out the constant in each of the terms.

$$2\int x^{3}\mathrm{d}x-\int x^{2}\mathrm{d}x+\int 4\mathrm{d}x$$

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}.

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

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

$$\frac{x^{4}}{2}-\frac{x^{3}}{3}+\int 4\mathrm{d}x$$

Find the integral of 4 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{x^{4}}{2}-\frac{x^{3}}{3}+4x$$

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.

$$\frac{6^{4}}{2}-\frac{6^{3}}{3}+4\times 6-\left(\frac{\left(-2\right)^{4}}{2}-\frac{\left(-2\right)^{3}}{3}+4\left(-2\right)\right)$$

Simplify.

$$\frac{1792}{3}$$