Animated Solution for Mathematics - Differentiation: Let $f(x)=\int e^x(x-1)(x-2)dx$. Then $f$ decreases in the interval

Imagine you are standing before a complex-looking function:

At first glance, it looks like a monster. It has an integral, an exponential term, and a quadratic expression all tangled together.

Many students would immediately try to solve the integral, diving headfirst into integration by parts, hoping to find the explicit form of $f(x)$. But here is the secret of the JEE Advanced: the problem is not asking you to find $f(x)$; it is asking you to understand its behavior.

Specifically, it wants to know where $f(x)$ is decreasing. In the world of calculus, 'decreasing' is a code word for 'negative slope'.

We do not need to integrate this function. We need to differentiate it.

This is the magic of the Fundamental Theorem of Calculus. It tells us that the derivative of an integral is simply the integrand itself.

If we define $f(x)=\int g(t)dt$, then $f^{\prime }(x)=g(x)$. By applying this, the integral sign vanishes, and we are left with the core of the problem:

Suddenly, the monster is tamed. We have a clear, simple expression for the slope of our function.

Now, we need to find where $f(x)$ is decreasing, which means we need to solve the inequality $f^{\prime }(x)<0$. Substituting our derivative, we get:

Here is where you must be sharp. Look at the term $e^x$. In the landscape of real numbers, $e^x$ is a constant guardian—it is always strictly positive.

It never touches the $x$-axis, and it never dips below it. Because it is always positive, we can divide both sides of our inequality by $e^x$ without worrying about the inequality sign flipping.

It is like dividing by a positive number like $5$ or $10$. Our inequality simplifies beautifully to:

We are now left with a classic quadratic inequality. We have two critical points: $x=1$ and $x=2$. These are the points where the slope is zero.

To find where the slope is negative, we use the Wavy Curve method. Imagine a number line.

To the right of $2$, if we pick a number like $3$, both $(x-1)$ and $(x-2)$ are positive, so their product is positive. Between $1$ and $2$, if we pick $1.5$, the first term is positive, but the second is negative, making the product negative.

To the left of $1$, both terms are negative, making the product positive again. We are looking for the region where the product is negative.

That region is exactly between $1$ and $2$. Thus, the function $f(x)$ is decreasing in the open interval $(1,2)$.

You have navigated the trap, used the theorem, and arrived at the truth. Calculus is not about memorizing formulas; it is about seeing the path through the complexity.