Solve ∫ (from 3 to 8) of 3x^2-6x+4 wrt x

$$\int _ { 3 } ^ { 8 } 3 x ^ { 2 } - 6 x + 4 d x$$

Evaluate

$$340$$

Solution Steps

Evaluate the indefinite integral first.

$$\int 3x^{2}-6x+4\mathrm{d}x$$

Integrate the sum term by term.

$$\int 3x^{2}\mathrm{d}x+\int -6x\mathrm{d}x+\int 4\mathrm{d}x$$

Factor out the constant in each of the terms.

$$3\int x^{2}\mathrm{d}x-6\int x\mathrm{d}x+\int 4\mathrm{d}x$$

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

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

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

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

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

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

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.

$$8^{3}-3\times 8^{2}+4\times 8-\left(3^{3}-3\times 3^{2}+4\times 3\right)$$

Simplify.

$$340$$