News: Currently the LaTeX and hidden solutions on this blog do not work on Google Reader.
Email me if you have suggestions on how to improve this blog!

Wednesday, 17 August 2011

Math GRE - #17

Let $h$ be the function defined by \[h(x)=\int_0^{x^2}{e^{x+t}dt}.\] Then $h'(1)=$

  • $e-1$
  • $e^2$
  • $e^2-e$
  • $2e^2$
  • $3e^2-e$

Solution :

The answer is $3e^2-e$.

This can be solved with applications of the product rule, the chain rule, and the fundamental theorem of calculus.

First let us recall that the derivative of an integral is itself, that is: \[\frac{d}{dx}\int_a^x{f(x)}=f(x).\]
When the upper limit is a function, we must apply the chain rule: \[\frac{d}{dx}\int_a^{x^2}{f(x)}=2x\cdot f(x^2).\]
For this particular question, we have: \[h(x)=\int_0^{x^2}{e^{x+t}dt}=e^x\cdot\int_0^{x^2}{e^{t}dt}=e^x\cdot g(x)\] for $g(x)=\int_0^{x^2}{e^{t}dt}$. The product rule tells us that \[h'(x)=e^x\cdot g(x)+e^x\cdot g'(x)=e^x\cdot\int_0^{x^2}{e^{t}dt}+e^x\cdot\left(2x\cdot e^{x^2}\right).\]
Upon substituting $1$ for $x$, we obtain the answer: \[h'(1)=e\cdot(e-1)+e\cdot(2\cdot e)=3e^2-e.\]


Post a Comment

This webpage is LaTeX enabled. To type in-line formulae, type your stuff between two '$'. To type centred formulae, type '\[' at the beginning of your formula and '\]' at the end.