Question

помогите решить интегралы​

Answer 1

Ответ:

$1)\ \ \int\limits^4_1 \sqrt{x}\Big(x-\dfrac{1}{x}\Big)\, dx=\int\limits^4_1\Big(x\sqrt{x}-\dfrac{1}{\sqrt{x}}\Big)\, dx=\int\limits^4_1\Big(x^{3/2}-x^{-1/2}\Big)\, dx=\\\\\\=\Big(\dfrac{x^{5/2}}{5/2}-\dfrac{x^{1/2}}{1/2}\Big)\Big|_1^4=\Big(\dfrac{2\sqrt{x^5}}{5}-2\sqrt{x}\Big)\Big|_1^4=\dfrac{2}{5}\sqrt{2^5}-2\sqrt{4}-\Big(\dfrac{2}{5}\sqrt{1^5}-2\sqrt{1}\Big)=\\\\\\=\dfrac{8\sqrt2}{5}-4-\dfrac{2}{5} +2=\dfrac{8\sqrt2}{5}-\dfrac{12}{5}=\dfrac{4\, (2\sqrt5-3)}{5}$

$2)\ \ \int\limits^8_14\sqrt[3]{x}\Big(1-\dfrac{4}{x}\Big)\, dx=\int\limits^8_1\Big(4x^{1/3}-\dfrac{16}{x^{2/3}}\Big)\, dx=\Big(\dfrac{4x^{4/3}}{4/3}-\dfrac{16x^{1/3}}{1/3}\Big)\Big|_1^8=\\\\\\=\Big(3\sqrt[3]{x^4}-48\sqrt[3]{x}\Big)\Big|_1^8=(3\cdot 8\sqrt[3]{8}-48\cdot 2\sqrt[3]{2})-(3-48)=\\\\=48\sqrt[3]{2}-96\sqrt[3]{2}+45=-48\sqrt[3]{2}+45$