Answer :

konrad509
[tex]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}=\\ [/tex]
[tex]\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 [/tex]
Yeah, you'd have to use the inverse function to produce this result.

Let's get the inverse function first:

[tex]y={ x }^{ 3 }-2\\ \\ { x }^{ 3 }=y+2\\ \\ x=\sqrt [ 3 ]{ y+2 }[/tex]

[tex]\\ \\ \therefore \quad { f }^{ -1 }\left( x \right) =\sqrt [ 3 ]{ x+2 } [/tex]

Now, we can solve the problem using:

[tex]\int _{ -1 }^{ 25 }{ \sqrt [ 3 ]{ x+2 } } dx[/tex]

But to solve the problem more easily we make u=x+2, therefore du/dx=1, therefore du=dx.

When x=25, u=27.

When x=-1, u=1.

Now:

[tex]\int _{ 1 }^{ 27 }{ { u }^{ \frac { 1 }{ 3 } } } du\\ \\ ={ \left[ \frac { 3 }{ 4 } { u }^{ \frac { 4 }{ 3 } } \right] }_{ 1 }^{ 27 }[/tex]

[tex]\\ \\ =\frac { 3 }{ 4 } \cdot { 27 }^{ \frac { 4 }{ 3 } }-\frac { 3 }{ 4 } \cdot { 1 }^{ \frac { 4 }{ 3 } }\\ \\ =\frac { 3 }{ 4 } { \left( { 3 }^{ 3 } \right) }^{ \frac { 4 }{ 3 } }-\frac { 3 }{ 4 }[/tex]

[tex]\\ \\ =\frac { 3 }{ 4 } \cdot { 3 }^{ 4 }-\frac { 3 }{ 4 } \\ \\ =\frac { 3 }{ 4 } \left( { 3 }^{ 4 }-1 \right)[/tex]

[tex]\\ \\ =\frac { 3 }{ 4 } \cdot 80\\ \\ =60[/tex]

Answer: 60 units squared.

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