Solve ∫ (from -3 to 4) of x^-(x^2-x-12) wrt x

$$\int_{ -3 }^{ 4 } { x }^{ } -( { x }^{ 2 } -x-12) d x$$

Evaluate

$$\frac{182}{3}\approx 60.666666667$$

Solution Steps

Calculate x to the power of 1 and get x.

$$\int _{-3}^{4}x-\left(x^{2}-x-12\right)\mathrm{d}x$$

To find the opposite of x^{2}-x-12, find the opposite of each term.

$$\int _{-3}^{4}x-x^{2}+x+12\mathrm{d}x$$

Combine x and x to get 2x.

$$\int _{-3}^{4}2x-x^{2}+12\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int 2x-x^{2}+12\mathrm{d}x$$

Integrate the sum term by term.

$$\int 2x\mathrm{d}x+\int -x^{2}\mathrm{d}x+\int 12\mathrm{d}x$$

Factor out the constant in each of the terms.

$$2\int x\mathrm{d}x-\int x^{2}\mathrm{d}x+\int 12\mathrm{d}x$$

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

$$x^{2}-\int x^{2}\mathrm{d}x+\int 12\mathrm{d}x$$

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}.

$$x^{2}-\frac{x^{3}}{3}+\int 12\mathrm{d}x$$

Find the integral of 12 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$x^{2}-\frac{x^{3}}{3}+12x$$

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.

$$4^{2}-\frac{4^{3}}{3}+12\times 4-\left(\left(-3\right)^{2}-\frac{\left(-3\right)^{3}}{3}+12\left(-3\right)\right)$$

Simplify.

$$\frac{182}{3}$$