Solve ∫ (from 1 / 2 to 4) of (x^3-6x^2+9x+1) wrt x

$$\int _ { 1 / 2 } ^ { 4 } ( x ^ { 3 } - 6 x ^ { 2 } + 9 x + 1 ) d x$$

Evaluate

$$\frac{679}{64}=10.609375$$

Solution Steps

Evaluate the indefinite integral first.

$$\int x^{3}-6x^{2}+9x+1\mathrm{d}x$$

Integrate the sum term by term.

$$\int x^{3}\mathrm{d}x+\int -6x^{2}\mathrm{d}x+\int 9x\mathrm{d}x+\int 1\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int x^{3}\mathrm{d}x-6\int x^{2}\mathrm{d}x+9\int x\mathrm{d}x+\int 1\mathrm{d}x$$

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}.

$$\frac{x^{4}}{4}-6\int x^{2}\mathrm{d}x+9\int x\mathrm{d}x+\int 1\mathrm{d}x$$

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 -6 times \frac{x^{3}}{3}.

$$\frac{x^{4}}{4}-2x^{3}+9\int x\mathrm{d}x+\int 1\mathrm{d}x$$

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

$$\frac{x^{4}}{4}-2x^{3}+\frac{9x^{2}}{2}+\int 1\mathrm{d}x$$

Find the integral of 1 using the table of common integrals rule \int a\mathrm{d}x=ax.

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

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.

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

Simplify.

$$\frac{679}{64}$$