Լուծեք ∫ (from - 1 to 1) of (2x^2-x^3) wrt x

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

Գնահատել

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

Լուծման քայլերը

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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}. Բազմապատկեք 2 անգամ \frac{x^{3}}{3}:

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

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}. Բազմապատկեք -1 անգամ \frac{x^{4}}{4}:

$$\frac{2x^{3}}{3}-\frac{x^{4}}{4}$$

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.

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

Պարզեցնել:

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