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.
