Լուծեք ∫ (from 0 to 2 pi) of int_0^1rsqrt{4r^2+1}drdθ

$$\int _ { 0 } ^ { 2 \pi } \int _ { 0 } ^ { 1 } r \sqrt { 4 r ^ { 2 } + 1 } d r d \theta$$

Գնահատել

$$\frac{5\pi \sqrt{5}-\pi }{6}\approx 5.3304135$$

Լուծման քայլերը

Evaluate the indefinite integral first.

$$\int \int _{0}^{1}r\sqrt{4r^{2}+1}\mathrm{d}r\mathrm{d}\theta$$

Find the integral of \int _{0}^{1}r\sqrt{4r^{2}+1}\mathrm{d}r using the table of common integrals rule \int a\mathrm{d}\theta =a\theta .

$$\int _{0}^{1}r\sqrt{4r^{2}+1}\mathrm{d}r\theta$$

Պարզեցնել:

$$\frac{5\sqrt{5}-1}{12}\theta$$

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.

$$\left(\frac{5}{12}\times 5^{\frac{1}{2}}-\frac{1}{12}\right)\times 2\pi -\left(\frac{5}{12}\times 5^{\frac{1}{2}}-\frac{1}{12}\right)\times 0$$

Պարզեցնել:

$$\frac{5\sqrt{5}\pi -\pi }{6}$$