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

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

Evaluate

$$\frac{4}{3}\approx 1.333333333$$

Solution Steps

Evaluate the indefinite integral first.

$$\int x^{3}-x^{2}+1\mathrm{d}x$$

Integrate the sum term by term.

$$\int x^{3}\mathrm{d}x+\int -x^{2}\mathrm{d}x+\int 1\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int x^{3}\mathrm{d}x-\int x^{2}\mathrm{d}x+\int 1\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{3}\mathrm{d}x with \frac{x^{4}}{4}.

$$\frac{x^{4}}{4}-\int x^{2}\mathrm{d}x+\int 1\mathrm{d}x$$

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 -1 times \frac{x^{3}}{3}.

$$\frac{x^{4}}{4}-\frac{x^{3}}{3}+\int 1\mathrm{d}x$$

Find the integral of 1 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{x^{4}}{4}-\frac{x^{3}}{3}+x$$

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.

$$\frac{1^{4}}{4}-\frac{1^{3}}{3}+1-\left(\frac{\left(-1\right)^{4}}{4}-\frac{\left(-1\right)^{3}}{3}-1\right)$$

Simplify.

$$\frac{4}{3}$$