Լուծեք ∫ (from 0 to 17) of (9x^2+48x^3)

$$\int _ { 0 } ^ { 17 } ( 9 x ^ { 2 } + 48 x ^ { 3 } )$$

Գնահատել

$$1016991$$

Լուծման քայլերը

Evaluate the indefinite integral first.

$$\int 9x^{2}+48x^{3}\mathrm{d}x$$

Integrate the sum term by term.

$$\int 9x^{2}\mathrm{d}x+\int 48x^{3}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$9\int x^{2}\mathrm{d}x+48\int x^{3}\mathrm{d}x$$

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 9 անգամ \frac{x^{3}}{3}:

$$3x^{3}+48\int x^{3}\mathrm{d}x$$

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 48 անգամ \frac{x^{4}}{4}:

$$3x^{3}+12x^{4}$$

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.

$$3\times 17^{3}+12\times 17^{4}-\left(3\times 0^{3}+12\times 0^{4}\right)$$

Պարզեցնել:

$$1016991$$