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

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

Evaluate

$$-2$$

Solution Steps

Use the distributive property to multiply 2x-3 by 4x^{2}+1.

$$\int _{0}^{2}8x^{3}+2x-12x^{2}-3\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int 8x^{3}+2x-12x^{2}-3\mathrm{d}x$$

Integrate the sum term by term.

$$\int 8x^{3}\mathrm{d}x+\int 2x\mathrm{d}x+\int -12x^{2}\mathrm{d}x+\int -3\mathrm{d}x$$

Factor out the constant in each of the terms.

$$8\int x^{3}\mathrm{d}x+2\int x\mathrm{d}x-12\int x^{2}\mathrm{d}x+\int -3\mathrm{d}x$$

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

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

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}.

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

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 -12 times \frac{x^{3}}{3}.

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

Find the integral of -3 using the table of common integrals rule \int a\mathrm{d}x=ax.

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

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.

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

Simplify.

$$-2$$