Question

Use part 1 of the Fundamental Theorem of Calculus to find the derivative of the function.

Answer 1

\[\int\limits_{2}^{1/x} \arctan t dt\]

Answer 2

Let
\[ F(t) = \int f(t) \ dt \]
Then by the FTC, \( dF/dt = f(t) \).

So use this with your problem. Let \( F(t) = \int \arctan t \ dt \).

Then
\[ \int_a^{1/x} \arctan t \ dt = F(1/x) - F(a) \]

Now differentiate with respect to x.

Answer 3

F'(1/x) (1/x^2)dx - F'(a) ? This is part 2 if I remember correctly

Answer 4

not quite.

Answer 5

Remember that F'(x) = arctan(x). Now use the chain rule carefully.

Notice also that F(a) is a number, a constant. Hence it's derivative is .....

Answer 6

I think I got the answer by \[h(x) = \int\limits_{2}^{1/x} \arctan t dt =- \arctan (1/x) *(1/x ^{2}) = -\arctan(1/x)/x ^{2}\]

Answer 7

be more careful with your notation. The integral is not equal to the expressions on the right.

Answer 8

but you have the right answer. Just write it carefully.

Answer 9

I got the answer but still don't fully get what's going on

Answer 10

Here's what is going on. Let g be the function defined by
\[ g(x) = \int_a^{1/x} \arctan x \ dx \]
We want to find dg/dx.

That is what you've written down.

Now there are a number of intermediate steps: let \( F(x) = \int \arctan x \ dx \), then
\[ g(x) = F(1/x) - F(a) \]
Therefore
\[ \begin{align}
\frac{dg}{dx} &= \frac{d\ }{dx} F(1/x) - \frac{d\ }{dx} F(a) \\
&= \frac{-1}{x^2} F'(1/x) - 0 \\
&= -\arctan(1/x)/x^2 \ \ \ \ \ \ \ \ \ \ \ \ \ \hbox{ by the FTC}
\end{align} \]

Answer 11

Thanks, the whole d/dx was throwing me for a loop.