Solve ∫ (from - 3 to 1) of (-3x^2-5x+9) wrt x=

$$\int _ { - 3 } ^ { 1 } ( - 3 x ^ { 2 } - 5 x + 9 ) d x =$$

Evaluate

$$28$$

Solution Steps

Evaluate the indefinite integral first.

$$\int -3x^{2}-5x+9\mathrm{d}x$$

Integrate the sum term by term.

$$\int -3x^{2}\mathrm{d}x+\int -5x\mathrm{d}x+\int 9\mathrm{d}x$$

Factor out the constant in each of the terms.

$$-3\int x^{2}\mathrm{d}x-5\int x\mathrm{d}x+\int 9\mathrm{d}x$$

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 -3 times \frac{x^{3}}{3}.

$$-x^{3}-5\int x\mathrm{d}x+\int 9\mathrm{d}x$$

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 -5 times \frac{x^{2}}{2}.

$$-x^{3}-\frac{5x^{2}}{2}+\int 9\mathrm{d}x$$

Find the integral of 9 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$-x^{3}-\frac{5x^{2}}{2}+9x$$

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.

$$-1^{3}-\frac{5}{2}\times 1^{2}+9\times 1-\left(-\left(-3\right)^{3}-\frac{5}{2}\left(-3\right)^{2}+9\left(-3\right)\right)$$

Simplify.

$$28$$