Animated Solution for Mathematics - Definite Integration: Let $f:R\to R$ be a differentiable function and $f(1)=4$. Then the value of ${\lim }_{x\to 1}{\int }_4^{f(x)}\frac{2t}{x-1}dt$ is

We are tasked with evaluating the limit:

The first rule of limits is to test the behavior of the expression as $x$ approaches the target value. When we substitute $x=1$, the denominator $x-1$ becomes $1-1=0$.

Now, consider the numerator. As $x\to 1$, the upper limit of our integral, $f(x)$, approaches $f(1)$. Given that $f(1)=4$, the integral becomes:

Geometrically, this represents the area under the curve $y=2t$ over an interval of zero width. We have confirmed the presence of a $\frac{0}{0}$ indeterminate form, which signals that we should apply L'Hopital's Rule.

L'Hopital's Rule states that for a $\frac{0}{0}$ form, the limit of the ratio is equal to the limit of the ratio of the derivatives. We must calculate the derivative of the integral with respect to $x$:

To solve this efficiently, we use the Newton-Leibniz Formula (the Chain Rule for Integrals). This rule states:

Applying this rule to our specific integral, we replace $t$ with $f(x)$ and multiply by the derivative $f^{\prime }(x)$. The derivative of the constant lower limit $4$ is zero, so it does not contribute to the result.

The derivative of the numerator is $2(f(x))\cdot f^{\prime }(x)$, and the derivative of the denominator $x-1$ is $1$. Our limit now simplifies to:

Since $f$ is differentiable, it is also continuous. We can evaluate the limit by direct substitution of $x=1$:

Substituting the known value $f(1)=4$, we obtain the final result: