Solve ∫ (from 1 to 3) of (6x^2+8x-5) wrt x=

$$\int _ { 1 } ^ { 3 } ( 6 x ^ { 2 } + 8 x - 5 ) d x =$$

Evaluate

$$74$$

Solution Steps

Evaluate the indefinite integral first.

$$\int 6x^{2}+8x-5\mathrm{d}x$$

Integrate the sum term by term.

$$\int 6x^{2}\mathrm{d}x+\int 8x\mathrm{d}x+\int -5\mathrm{d}x$$

Factor out the constant in each of the terms.

$$6\int x^{2}\mathrm{d}x+8\int x\mathrm{d}x+\int -5\mathrm{d}x$$

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}.

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

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 8 times \frac{x^{2}}{2}.

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

Find the integral of -5 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$2x^{3}+4x^{2}-5x$$

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.

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

Simplify.

$$74$$