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

$$\int_{ -1 }^{ 2 } { x }^{ 3 } - { x }^{ 2 } d x$$

Գնահատել

$$\frac{3}{4}=0.75$$

Լուծման քայլերը

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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

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

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}. Բազմապատկեք -1 անգամ \frac{x^{3}}{3}:

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

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.

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

Պարզեցնել:

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