Question

Задание приложено...

Answer 1

Ответ:

1) $\boxed{ \boldsymbol{ \displaystyle \int\limits^{2}_{0} \, dx \int\limits^{1}_{0} {(x^{2} + 2y)} \, dy = \frac{14}{3} } }$

2) $\boxed{ \boldsymbol{ \displaystyle \int\limits^{2}_{1} \, dx \int\limits^{x}_{1} {\frac{x^{2}}{y^{2}} } \, dy = \frac{5}{6} } }$

3) Согласно условию: $\boxed{ \boldsymbol{ \displaystyle \int\limits^{2\pi }_{0} \, dx \int\limits^{a}_{a \sin \phi} {r} \, dr = \pi a^{2}\cos^{2} \phi } }$

Если условие содержит ошибку:

$\boxed{ \boldsymbol{ \displaystyle \int\limits^{2\pi }_{0} \, d\phi \int\limits^{a}_{a \sin \phi} {r} \, dr = \frac{\pi a^{2}}{2} } }$

Примечание:

По таблице интегралов:

$\boxed{\displaystyle \int x^{n} \ dx = \frac{x^{n + 1}}{n + 1} + C; n \neq -1, x > 0}$

$\boxed{\displaystyle \int \cos x \ dx = \sin x + C}$

Объяснение:

3.

1)

$\displaystyle \int\limits^{2}_{0} \, dx \int\limits^{1}_{0} {(x^{2} + 2y)} \, dy = \int\limits^{2}_{0} \Bigg( \bigg(yx^{2} + 2 \cdot \frac{y^{2}}{2} \bigg) \bigg|^{1}_{0} \Bigg) \, dx = \int\limits^{2}_{0} \Bigg( \bigg(yx^{2} + y^{2} \bigg) \bigg|^{1}_{0} \Bigg) \, dx =$

$\displaystyle = \int\limits^{2}_{0} \Bigg ( \bigg(1 \cdot x^{2} + 1^{2} \bigg) - \bigg(0 \cdot x^{2} + 0^{2} \bigg) \Bigg) \, dx = \int\limits^{2}_{0} \bigg(x^{2} + 1 \bigg) \, dx = \Bigg( \frac{x^{3}}{3} + x \Bigg) \Bigg|^{2}_{0} =$

$\displaystyle = \bigg( \frac{2^{3}}{3} + 2 \bigg) - \bigg( \frac{0^{3}}{3} + 0 \bigg) = \frac{8}{3} + 2 = \frac{8 + 6}{3} = \frac{14}{3}$

2)

$\displaystyle \int\limits^{2}_{1} \, dx \int\limits^{x}_{1} {\frac{x^{2}}{y^{2}} } \, dy = \int\limits^{2}_{1} x^{2} \, dx \int\limits^{x}_{1} { y^{-2} } \, dy = \int\limits^{2}_{1} \bigg( \frac{x^{2}y^{-2 + 1}}{-2 + 1} \bigg) \bigg |^{x}_{1} \, dx = - \int\limits^{2}_{1} \bigg(\frac{ x^{2}}{y} \bigg) \bigg |^{x}_{1} \, dx =$

$\displaystyle = - \int\limits^{2}_{1} \Bigg ( \bigg(\frac{ x^{2}}{x} \bigg) -\bigg(\frac{ x^{2}}{1} \bigg) \Bigg ) \, dx = - \int\limits^{2}_{1} \bigg(x - x^{2} \bigg) \, dx = \int\limits^{2}_{1} \bigg(x^{2} -x \bigg) \, dx=$

$\displaystyle = \bigg(\frac{x^{3}}{3} - \frac{x^{2}}{2} \bigg) \bigg|^{2}_{1} = \bigg(\frac{2^{3}}{3} - \frac{2^{2}}{2} \bigg) - \bigg(\frac{1^{3}}{3} - \frac{1^{2}}{2} \bigg) = \bigg(\frac{8}{3} -2 \bigg) - \bigg(\frac{1}{3} - \frac{1^}{2} \bigg) =$

$\displaystyle = \frac{8}{3} -2 - \frac{1}{3} + \frac{1}{2} = \frac{8 -1}{3} - 2 + \frac{1}{2} = \frac{7}{3} - 2 + \frac{1}{2} = \frac{14 - 12 + 3}{6} = \frac{5}{6}$

3)

$\displaystyle \int\limits^{2\pi }_{0} \, dx \int\limits^{a}_{a \sin \phi} {r} \, dr = \int\limits^{2\pi }_{0} \Bigg( \bigg( \frac{r^{2}}{2} \bigg) \bigg|^{a}_{a \sin \phi} \Bigg) \, dx = \frac{1}{2} \int\limits^{2\pi }_{0} } \bigg( a^{2} - a^{2} \sin^{2} \phi\bigg) \, dx =$

$\displaystyle = \frac{a^{2}}{2} \int\limits^{2\pi }_{0} } \bigg( 1 - \sin^{2} \phi\bigg) \, dx = \frac{a^{2}}{2} \int\limits^{2\pi }_{0} } \cos^{2} \phi \, dx = \frac{a^{2}\cos^{2} \phi}{2} \int\limits^{2\pi }_{0} } \, dx = \frac{a^{2}\cos^{2} \phi}{2} \cdot x \bigg|^{2\pi}_{0}=$

$\displaystyle = \frac{a^{2}\cos^{2} \phi}{2} \bigg(2\pi - 0 \bigg) = \frac{2\pi a^{2}\cos^{2} \phi}{2} = \pi a^{2}\cos^{2} \phi$

Если предположить, что в третьем номере опечатка, то условие будет выглядеть так:

$\displaystyle \int\limits^{2\pi }_{0} \, \boldsymbol{ d\phi} \int\limits^{a}_{a \sin \phi} {r} \, dr =$ (воспользуемся решением интеграла, который дан по условию, см. выше)

$\displaystyle = \frac{a^{2}}{2} \int\limits^{2\pi }_{0} } \cos^{2} \phi \, d\phi = \frac{a^{2}}{2} \int\limits^{2\pi }_{0} } \bigg( \frac{1 + \cos 2\phi}{2} \bigg)\, d\phi = \frac{a^{2}}{4} \int\limits^{2\pi }_{0} } \bigg( 1 + \cos 2\phi \bigg)\, d\phi =$

$\displaystyle = \frac{a^{2}}{4} \Bigg( \int\limits^{2\pi }_{0} } 1 \, d\phi + \frac{1}{2} \int\limits^{2\pi }_{0} } \cos 2\phi\, d(2\phi) \Bigg ) = \frac{a^{2}}{4} \Bigg( \phi \bigg |^{2\pi }_{0} + \frac{1}{2} \cdot \sin 2 \phi \bigg |^{2\pi }_{0} \Bigg) =$

$\displaystyle = \frac{a^{2}}{4} \Bigg( (2 \pi - 0) + \frac{1}{2} \bigg( \sin 4 \pi - \sin 0 \bigg) \Bigg) = \frac{a^{2}}{4} \Bigg( 2 \pi + \frac{1}{2} \bigg(0 - 0 \bigg) \Bigg) = \frac{2 \pi a^{2}}{4} = \frac{\pi a^{2}}{2}$