Animated Solution for Mathematics - Definite Integration: Determine a positive integer $n\le 5$, such that ${\int }_0^1{e}^x(x-1{)}^{n}dx=16-6e$

When you see an expression like $I_n={\int }_0^1{e}^x(x-1{)}^{n}dx$, your first instinct might be panic. You might think about expanding the term or using complex substitutions.

However, in JEE Advanced, complexity is often a mask for a beautiful, underlying simplicity. We are looking for a positive integer $n\le 5$ such that this integral equals $16-6e$.

We define our target as $I_n$. To solve this, we use the method of Integration by Parts, defined by the formula:

Let $u=(x-1{)}^n$ and $dv=e^{x}dx$. Differentiating $u$ gives $du=n(x-1{)}^{n-1}dx$, and integrating $dv$ gives $v=e^x$.

Substituting these into our formula, we obtain:

Look at the boundary term! At $x=1$, $(1-1{)}^n=0$. At $x=0$, we get $-(0-1{)}^n{e}^0=-(-1{)}^n=(-1{)}^{n+1}$.

This simplifies our expression significantly, leading to the recurrence relation:

Now, we establish our foundation. For $n=0$:

With this base case, we iterate to find the value of $n$:

For $n=1$:

For $n=2$:

For $n=3$:

We have arrived at the target value. The condition is satisfied when $n=3$.

This is the power of the reduction method—we did not brute force the integral; we built a mathematical ladder and climbed it to the solution.