Пожалуйста решите номер 33 обе буквы и "а" и "б"

Ответ:

$1)\ \ \int \dfrac{dx}{\sqrt{x}+\sqrt[4]{x}}=\Big[\ x=t^4\ ,\ dx=4t^3\, dt\ \Big]=\left \int \right .\dfrac{4t^3\, dt}{t^2+t}=4\int \dfrac{t^2\, dt}{t+1}=\\\\\\=4\int \Big(t-1+\dfrac{1}{t+1}\Big)\, dt=4\Big(\dfrac{t^2}{2}-t+ln|t+1|\Big)+C=\\\\\\=2\sqrt{x}-4\sqrt[4]{x}+4\, ln\Big|\sqrt[4]{x}+1\Big|+C$

$2)\ \ y=2-x^2\ ,\ \ y=0\\\\2-x^2=0\ \ ,\ \ (\sqrt2-x)(\sqrt2+x)=0\ \ ,\ \ x_1=-\sqrt2\ ,\ x_2=\sqrt2\\\\\\V=\pi \int\limits^a_b\, f^2(x)\, dx=\pi \int\limits^{\sqrt2}_{-\sqrt2}\, (2-x^2)^2\, dx=2\pi \int\limits^{\sqrt2}_{0}\, (2-x^2)^2\, dx=\\\\\\=2\pi \int\limits^{\sqrt2}_{0}\, (4-4x^2+x^4)\, dx=2\pi \cdot \Big(4x-\dfrac{4x^3}{3}+\dfrac{x^5}{5}\Big)\Big|_0^{\sqrt2}=\\\\\\=2\pi \cdot \Big(4\sqrt2-\dfrac{8\sqrt2}{3}+\dfrac{4\sqrt2}{5}\Big)=2\pi \cdot \dfrac{32\sqrt2}{15}=\dfrac{64\sqrt2}{15}$

Замена переменной:

$\sqrt[4]{x} =t\\\\ x=t^4\\\\dx=4t^3dt\\\\\\\sqrt{x} =t^2$

$\int{\frac{dx}{\sqrt{x} +\sqrt[4]{x} } \, =\int{\frac{4t^3dt}{t^2 +t } \, =4\int{\frac{t^2}{t +1 } \, dt=4\int{\frac{t^2-1+1}{t +1 } \, dt=4\int{\frac{t^2-1}{t +1 } \, dt+4\int{\frac{1}{t +1 } \, dt=$

$=4\int{(t -1) } \, dt+4\int{\frac{1}{t +1 } \, dt=4\frac{t^2}{2} -4t+4ln|t+1|+C=$

Обратная замена

$=2\sqrt{x} -4\sqrt[4]{x} +4ln|\sqrt[4]{x} +1|+C$

б)

$V_{Ox}=\pi \int\limits^a_b {f^2(x)} \, dx=\pi \int\limits^{\sqrt{2}}_{-\sqrt{2}} {(2-x^2)^2} \, dx=\pi \int\limits^{\sqrt{2}}_{-\sqrt{2}} {(4-4x^2+x^4)} \, dx=\\\\=\pi (4x-4\frac{x^3}{3}+\frac{x^5}{5} )|^{\sqrt{2}}_{-\sqrt{2}}=\frac{64\pi\sqrt{2} }{15}$