Solve ∫ (from - 1 to 2) of (x+1)(x-2)(x-4) wrt x

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

Evaluate

$$\frac{63}{4}=15.75$$

Solution Steps

Apply the distributive property by multiplying each term of x+1 by each term of x-2.

$$\int _{-1}^{2}\left(x^{2}-2x+x-2\right)\left(x-4\right)\mathrm{d}x$$

Combine -2x and x to get -x.

$$\int _{-1}^{2}\left(x^{2}-x-2\right)\left(x-4\right)\mathrm{d}x$$

Apply the distributive property by multiplying each term of x^{2}-x-2 by each term of x-4.

$$\int _{-1}^{2}x^{3}-4x^{2}-x^{2}+4x-2x+8\mathrm{d}x$$

Combine -4x^{2} and -x^{2} to get -5x^{2}.

$$\int _{-1}^{2}x^{3}-5x^{2}+4x-2x+8\mathrm{d}x$$

Combine 4x and -2x to get 2x.

$$\int _{-1}^{2}x^{3}-5x^{2}+2x+8\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int x^{3}-5x^{2}+2x+8\mathrm{d}x$$

Integrate the sum term by term.

$$\int x^{3}\mathrm{d}x+\int -5x^{2}\mathrm{d}x+\int 2x\mathrm{d}x+\int 8\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int x^{3}\mathrm{d}x-5\int x^{2}\mathrm{d}x+2\int x\mathrm{d}x+\int 8\mathrm{d}x$$

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}.

$$\frac{x^{4}}{4}-5\int x^{2}\mathrm{d}x+2\int x\mathrm{d}x+\int 8\mathrm{d}x$$

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 -5 times \frac{x^{3}}{3}.

$$\frac{x^{4}}{4}-\frac{5x^{3}}{3}+2\int x\mathrm{d}x+\int 8\mathrm{d}x$$

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}.

$$\frac{x^{4}}{4}-\frac{5x^{3}}{3}+x^{2}+\int 8\mathrm{d}x$$

Find the integral of 8 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{x^{4}}{4}-\frac{5x^{3}}{3}+x^{2}+8x$$

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.

$$\frac{2^{4}}{4}-\frac{5}{3}\times 2^{3}+2^{2}+8\times 2-\left(\frac{\left(-1\right)^{4}}{4}-\frac{5}{3}\left(-1\right)^{3}+\left(-1\right)^{2}+8\left(-1\right)\right)$$

Simplify.

$$\frac{63}{4}$$