![y=x^3-2\\ x^3=y+2\\ x=\sqrt[3]{y+2}\\\\ \int \limits_{-1}^{25}\sqrt[3]{y+2}\, dy=\\ \int \limits_{-1}^{25}(y+2)^{\tfrac{1}{3}}\, dy=\\ \left[\dfrac{(y+2)^{\tfrac{4}{3}}}{\frac{4}{3}} \right]_{-1}^{25}=\\ ](/tpl/images/0195/2031/a8849.png)
![\left[\dfrac{3}{4}(y+2)^{\tfrac{4}{3}} \right]_{-1}^{25}=\\ \left[\dfrac{3}{4}(y+2)\sqrt[3]{y+2} \right]_{-1}^{25}=\\ \dfrac{3}{4}(25+2)\sqrt[3]{25+2}-\left(\dfrac{3}{4}(-1+2)\sqrt[3]{-1+2}\right)=\\ \dfrac{3}{4}\cdot27\sqrt[3]{27}-\left(\dfrac{3}{4}\sqrt[3]{1}\right)=\\ \dfrac{3}{4}\cdot27\cdot3-\dfrac{3}{4}=\\ \dfrac{3}{4}(81-1)=\\ \dfrac{3}{4}\cdot 80=\\3\cdot20=\\ 60 ](/tpl/images/0195/2031/59a16.png)
![y={ x }^{ 3 }-2\\ \\ { x }^{ 3 }=y+2\\ \\ x=\sqrt [ 3 ]{ y+2 }](/tpl/images/0195/2031/221ef.png)
![\\ \\ \therefore \quad { f }^{ -1 }\left( x \right) =\sqrt [ 3 ]{ x+2 }](/tpl/images/0195/2031/d5a32.png)
![\int _{ -1 }^{ 25 }{ \sqrt [ 3 ]{ x+2 } } dx](/tpl/images/0195/2031/4802d.png)
![\int _{ 1 }^{ 27 }{ { u }^{ \frac { 1 }{ 3 } } } du\\ \\ ={ \left[ \frac { 3 }{ 4 } { u }^{ \frac { 4 }{ 3 } } \right] }_{ 1 }^{ 27 }](/tpl/images/0195/2031/64aba.png)
Ex 3.7
12. find the area between the curve y=x³-2 and the y-axis between y= -1 and y=25...
Mathematics, 25.08.2019 01:30 pineapplepizaaaaa
Ex 3.7
12. find the area between the curve y=x³-2 and the y-axis between y= -1 and y=25
Answers: 2
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