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Forums
Mathematics
Calculus
Integral with an inverse function limit
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[QUOTE="John Park, post: 5699337, member: 616032"] Let me struggle through this in my own way. I find it useful to give familiar-looking names to things wherever possible. So let's put [I]z[/I][SUP]-1[/SUP]([I]c[/I]) = [I]y[/I] and [I]∫z(x)[/I]d[I]x = Z[/I], and I'll call the integral-derivative we have to evaluate[I] F. [/I] So we have to evaluate [I]F [/I]= (d/d[I]c[/I])( [I]Z[/I](1) -[I]Z[/I]([I]y[/I]) ), which immediately becomes -(d/d[I]c[/I])[I]Z[/I]([I]y[/I]) = -(d[I]Z[/I]([I]y)[/I]/d[I]y[/I]). (d[I]y[/I]/d[I]c[/I]) = [I]z[/I]([I]y[/I]).(d[I]y[/I]/d[I]c[/I]) But [I]z[/I]([I]y[/I]) = [I]z[/I]([I]z[/I][SUP]-1[/SUP]([I]c[/I])) = [I]c[/I] So we have [I]F [/I]= [I]c[/I].(dy/d[I]c[/I]) . . . . . . .(A) Now [I]z[/I][SUP]-1[/SUP]([I]c[/I]) = [I]y[/I] by definition; so [I]c [/I]= z([I]y[/I]), and, differentiating, d[I]c[/I]/d[I]y[/I] =d[I]z[/I]/d[I]y[/I], or if we invert, d[I]y[/I]/d[I]c [/I]= 1/(d[I]z[/I]/d[I]y[/I]) = 1/[I]z[/I]'([I]y[/I]) = 1/[I]z[/I]'( [I]z[/I][SUP]-1[/SUP]([I]c[/I]) ) Substitute this for d[I]y[/I]/d[I]c[/I] in (A) and I think we're done. [/QUOTE]
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Forums
Mathematics
Calculus
Integral with an inverse function limit
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