Animated Solution for Mathematics - Definite Integration: The area enclosed between the curve $y={\log }_e(x+e)$ and the coordinate axes is

Welcome, fellow traveler on the JEE journey. Today, we are going to demystify a beautiful problem involving the natural logarithm. Many students fear logarithmic curves, but once you understand their geometry, they become some of the most elegant shapes in calculus.

Let us look at the curve $y={\log }_e(x+e)$.

Imagine the standard natural log curve, $y={\log }_e(x)$. It hugs the y-axis, passing through $(1,0)$ and climbing slowly toward infinity.

Now, look at our function: $y={\log }_e(x+e)$. That $+e$ inside the argument is a transformation. In coordinate geometry, replacing $x$ with $x+h$ shifts the graph to the left by $h$ units.

So, our curve is simply the standard log curve shifted left by $e$ units. The vertical asymptote, which was at $x=0$, has now migrated to $x=-e$. This visualization is your first step to mastery.

To find the area enclosed between the curve and the coordinate axes, we need to know exactly where the curve 'hits' the axes. These are our boundaries.

First, the y-intercept occurs where $x=0$. Plugging this into our equation, we get $y={\log }_e(0+e)={\log }_e(e)$. Since the logarithm of a base to itself is always $1$, our curve crosses the y-axis at $(0,1)$.

Next, the x-intercept occurs where $y=0$. Setting the equation to zero, we have $0={\log }_e(x+e)$.

To solve for $x$, we exponentiate both sides: $e^0=x+e$. Since $e^0=1$, we get $1=x+e$, which simplifies to $x=1-e$. So, the curve crosses the x-axis at $(1-e,0)$.

We now have our region: it is bounded by the x-axis from $x=1-e$ to $x=0$, and by the curve $y={\log }_e(x+e)$ from above.

Now, we set up the integral for the area $A$:

This looks a bit intimidating, but here is where we use the 'JEE Toolkit'. We can simplify this using a substitution. Let $t=x+e$. Then, $dt=dx$.

We must be careful with the limits! When $x=1-e$, $t=(1-e)+e=1$. When $x=0$, $t=0+e=e$.

Our integral transforms into something much friendlier:

We are now looking at the integral of ${\log }_e(t)$. This is a standard result that every JEE aspirant should have in their mental library. Using integration by parts, we know that $\int {\log }_e(t)dt=t{\log }_e(t)-t$.

Now, we apply our limits from $1$ to $e$:

Let's evaluate this step-by-step. First, the upper limit $t=e$:

Next, the lower limit $t=1$:

Finally, we subtract the lower limit value from the upper limit value:

The area enclosed is exactly $1$ square unit. It is a beautiful, clean integer result.

This problem teaches us that even when a function looks shifted or complex, a simple substitution can reveal the underlying simplicity. Keep this logic in your arsenal, and you will tackle any area-under-the-curve problem with confidence. Happy solving!