Solve ∫ (from 0 to 1) of (3x^3-x^2+2x-4) wrt x

$$\int_{ 0 }^{ 1 } (3 { x }^{ 3 } - { x }^{ 2 } +2x-4) d x$$

Evaluate

$$-\frac{31}{12}\approx -2.583333333$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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

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

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}.

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

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}.

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

Find the integral of -4 using the table of common integrals rule \int a\mathrm{d}x=ax.

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

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.

$$\frac{3}{4}\times 1^{4}-\frac{1^{3}}{3}+1^{2}-4-\left(\frac{3}{4}\times 0^{4}-\frac{0^{3}}{3}+0^{2}-4\times 0\right)$$

Simplify.

$$-\frac{31}{12}$$