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

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

Evaluate

$$1$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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}.

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

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

$$\frac{x^{3}+2x^{3}}{3}$$

Simplify.

$$x^{3}$$

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.

$$1^{3}-0^{3}$$

Simplify.

$$1$$