Solve ∫ (from 0 to 1) of (2x+3)*(3x-5) wrt x

$$\int _ { 0 } ^ { 1 } ( 2 x + 3 ) \cdot ( 3 x - 5 ) d x$$

Evaluate

$$-\frac{27}{2}=-13.5$$

Solution Steps

Apply the distributive property by multiplying each term of 2x+3 by each term of 3x-5.

$$\int _{0}^{1}6x^{2}-10x+9x-15\mathrm{d}x$$

Combine -10x and 9x to get -x.

$$\int _{0}^{1}6x^{2}-x-15\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int 6x^{2}-x-15\mathrm{d}x$$

Integrate the sum term by term.

$$\int 6x^{2}\mathrm{d}x+\int -x\mathrm{d}x+\int -15\mathrm{d}x$$

Factor out the constant in each of the terms.

$$6\int x^{2}\mathrm{d}x-\int x\mathrm{d}x+\int -15\mathrm{d}x$$

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}.

$$2x^{3}-\int x\mathrm{d}x+\int -15\mathrm{d}x$$

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 -1 times \frac{x^{2}}{2}.

$$2x^{3}-\frac{x^{2}}{2}+\int -15\mathrm{d}x$$

Find the integral of -15 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$2x^{3}-\frac{x^{2}}{2}-15x$$

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.

$$2\times 1^{3}-\frac{1^{2}}{2}-15-\left(2\times 0^{3}-\frac{0^{2}}{2}-15\times 0\right)$$

Simplify.

$$-\frac{27}{2}$$