Question

How do you find $F'\left( x \right)$ given $F\left( x \right) = \smallint \dfrac{1}{t}dt$ from $\left[ {1,\,{x^2}} \right]$ ?

Answer 1

Here we have to find the derivatives of the given term and use the given limits.
Integration is the algebraic method of finding the integral for a function at any point on the graph. On finding the integral of a function with respect to $x$ means finding the area to the $x$ axis from the curve
The integral is usually called the anti-derivatives, because integrating is the reverse process of differentiating. If asked to find the derivative of an integral using the fundamental theorem of calculus, integrate the function and then apply the limits and solve..

Complete step by step answer:
The given function is $F\left( x \right) = \smallint \dfrac{1}{t}dt$
We know that by the theorem of calculus
$\Rightarrow \smallint \dfrac{1}{x}dx = \ln \left( x \right) + c$
By applying the above equation in $F\left( x \right)$ , we get
$\Rightarrow F\left( x \right) = \smallint \dfrac{1}{t}dt = \left[ {\ln t} \right]$
Now by applying the limits $\left[ {1,\,{x^2}} \right]$
We get,
$\Rightarrow F\left( x \right) = \left[ {\ln t} \right]_1^{{x^2}}$
Now substituting the limits we get
$\Rightarrow F\left( x \right) = \left[ {\ln {x^2} - \ln 1} \right]$
In the above equation $\ln \left( 1 \right) = 0$
Therefore, we get
$\Rightarrow F\left( x \right) = \left[ {\ln {x^2}} \right]$
That can be written as
$\Rightarrow F\left( x \right) = 2\ln x$
By the fundamental theorem of calculus, we get: $\dfrac{d}{{dx}}\left( {\ln x} \right) = \dfrac{1}{x}\,for\,x > 0$
From that and the chain rule, we also get $\dfrac{d}{{dx}}\left( {\ln \left( { - x} \right)} \right) = \dfrac{1}{x}\,for\,x < 0$
On an interval that excludes 0, the anti-derivative of $\dfrac{1}{x}$ is $\ln x$ if the interval consists of positive numbers and it is $\ln \left( { - x} \right)$ if the interval consists of negative numbers.
$\therefore \ln |x|$ covers both cases.

Therefore $F'\left( x \right) = \left[ {2\ln x} \right] = 2 \times \dfrac{1}{x} = \dfrac{2}{x}$ .

Note: The term integral may also refer to the notion of the anti- derivative, a function $F$ whose Derivative is the given function $f$. In this case, it is called an indefinite integral and is
Written as, $\int {f\left( x \right)} dx = F\left( x \right) + C$. Integration is linear, additive, and preserves inequality of functions. The definite integral of $f$ over the interval $a$ to $b$ is given by $\int_a^b f = F|_a^b$ where $F$ is an anti- derivative of $f$. Integration is the operation of finding the region in the x-y plane bound by the function.