Solve ∫ (from 0 to 1) of (8x+3x^2) wrt x=

$$\int _ { 0 } ^ { 1 } ( 8 x + 3 x ^ { 2 } ) d x =$$

Evaluate

$$5$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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}.

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

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}.

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

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.

$$4\times 1^{2}+1^{3}-\left(4\times 0^{2}+0^{3}\right)$$

Simplify.

$$5$$