Լուծեք ∫ (from 0 to 10) of 2x+8 wrt x

$$\int_{ 0 }^{ 10 } 2x+8 d x$$

Գնահատել

$$180$$

Լուծման քայլերը

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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

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

Find the integral of 8 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$x^{2}+8x$$

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.

$$10^{2}+8\times 10-\left(0^{2}+8\times 0\right)$$

Պարզեցնել:

$$180$$