Solve ∫ (from - 1 to 2) of (4x^2-24x+9)

$$\int _ { - 1 } ^ { 2 } ( 4 x ^ { 2 } - 24 x + 9 )$$

Evaluate

$$3$$

Solution Steps

Evaluate the indefinite integral first.

$$\int 4x^{2}-24x+9\mathrm{d}x$$

Integrate the sum term by term.

$$\int 4x^{2}\mathrm{d}x+\int -24x\mathrm{d}x+\int 9\mathrm{d}x$$

Factor out the constant in each of the terms.

$$4\int x^{2}\mathrm{d}x-24\int x\mathrm{d}x+\int 9\mathrm{d}x$$

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

$$\frac{4x^{3}}{3}-24\int x\mathrm{d}x+\int 9\mathrm{d}x$$

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 -24 times \frac{x^{2}}{2}.

$$\frac{4x^{3}}{3}-12x^{2}+\int 9\mathrm{d}x$$

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

$$\frac{4x^{3}}{3}-12x^{2}+9x$$

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.

$$\frac{4}{3}\times 2^{3}-12\times 2^{2}+9\times 2-\left(\frac{4}{3}\left(-1\right)^{3}-12\left(-1\right)^{2}+9\left(-1\right)\right)$$

Simplify.

$$3$$