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

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

Evaluate

$$6$$

Solution Steps

Evaluate the indefinite integral first.

$$\int 4x^{3}-6x\mathrm{d}x$$

Integrate the sum term by term.

$$\int 4x^{3}\mathrm{d}x+\int -6x\mathrm{d}x$$

Factor out the constant in each of the terms.

$$4\int x^{3}\mathrm{d}x-6\int x\mathrm{d}x$$

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

$$x^{4}-6\int x\mathrm{d}x$$

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 -6 times \frac{x^{2}}{2}.

$$x^{4}-3x^{2}$$

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.

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

Simplify.

$$6$$