Question

A given function $y=f\left( x \right)$ satisfies (i) $f\left( 0 \right)=0$ (ii) ${{f}^{''}}\left( x \right)={{f}^{'}}\left( x \right)$ (iii) ${{f}^{'}}\left( 0 \right)=1$. Then find the area bounded by the graph $y=f\left( x \right)$ and the lines $x=0,x-1=0$ and $y+1=0$.
A. $e$ B. $e-2$ C. $e-1$ D. $e+1$

Answer 1

Hint: We use the given value to try to find out the function. Using methods of integration, we find out the values of constant parts in the integration. After we find y we put all the given functions in the graph to find out the area bounded by those graphs. Then we find the area of the curve using the intersecting points. Breaking the area into parts we find the area.

Complete step by step answer:
First, we need to find out the curve of $y=f\left( x \right)$.
We are given that $f\left( 0 \right)=0$, ${{f}^{''}}\left( x \right)={{f}^{'}}\left( x \right)$, ${{f}^{'}}\left( 0 \right)=1$.
We take the relation ${{f}^{''}}\left( x \right)={{f}^{'}}\left( x \right)$. We know that ${{f}^{''}}\left( x \right)=\dfrac{d}{dx}\left[ {{f}^{'}}\left( x \right) \right]$.
So, changing the equation we get ${{f}^{''}}\left( x \right)=\dfrac{d}{dx}\left[ {{f}^{'}}\left( x \right) \right]={{f}^{'}}\left( x \right)$.
We rearrange the equation to get the differential form
$\begin{align}
  & \dfrac{d}{dx}\left[ {{f}^{'}}\left( x \right) \right]={{f}^{'}}\left( x \right) \\
 & \Rightarrow \dfrac{d\left( {{f}^{'}}\left( x \right) \right)}{{{f}^{'}}\left( x \right)}=dx \\
\end{align}$

Now, we integrate both sides w.r.t. to get $\int{\dfrac{d\left( {{f}^{'}}\left( x \right) \right)}{{{f}^{'}}\left( x \right)}}=\int{dx}+c$.
The solution of the integration is $\log \left| {{f}^{'}}\left( x \right) \right|=x+c$.
In the equation we put x = 0. We get $\log \left| {{f}^{'}}\left( 0 \right) \right|=0+c=c$.
We know ${{f}^{'}}\left( 0 \right)=1$. So, $c=\log \left| 1 \right|=0$.
The equation becomes $\log \left| {{f}^{'}}\left( x \right) \right|=x$ which implies ${{f}^{'}}\left( x \right)={{e}^{x}}$.
We know ${{f}^{'}}\left( x \right)={{e}^{x}}\Rightarrow \dfrac{d\left( f\left( x \right) \right)}{dx}={{e}^{x}}$.
Now we integrate the equation ${{f}^{'}}\left( x \right)={{e}^{x}}$ w.r.t x again.
So, \[\int{d\left( f\left( x \right) \right)}=\int{d\left( {{e}^{x}} \right)}+C\] which means \[f\left( x \right)={{e}^{x}}+C\].
We have $f\left( 0 \right)=0$. Putting the value, x = 0 we get \[f\left( 0 \right)={{e}^{0}}+C\Rightarrow 0=1+C\Rightarrow C=-1\].
The equation becomes \[y=f\left( x \right)={{e}^{x}}-1\].
Now, we need to find the area bounded by the graph $y=f\left( x \right)={{e}^{x}}-1$ and the lines $x=0,x-1=0$ and $y+1=0$.
We draw all the given graphs to find the area which is marked.

We first find the yellow marked area as it is a square part with side length 1 unit.
So, area will be $1\times 1=1$ sq. unit.
Now we find the red marked area with the help of integration.
The area is under the curve $y=f\left( x \right)={{e}^{x}}-1$ from x = 0 to x = 1.
So, the integration becomes $\int\limits_{0}^{1}{ydx}$. We put the function to get $\int\limits_{0}^{1}{\left( {{e}^{x}}-1 \right)dx}$.
The area is \[\int\limits_{0}^{1}{\left( {{e}^{x}}-1 \right)dx}=\left[ {{e}^{x}}-x \right]_{0}^{1}=\left[ {{e}^{1}}-{{e}^{0}} \right]-\left[ 1-0 \right]=e-1-1=e-2\] sq. unit.
Total area becomes $e-2+1=e-1$ sq. unit. Correct option is (C).

Note:
We can’t take the whole area covered by the curve as it has to be bounded by $x=0,x-1=0$. Although the yellow marked area is under the curve, it is on the negative side of the x-axis. So, if we integrate the whole thing it will cancel out the positive part’s area because of its sign. The area can’t be negative. so, we have to find out the areas separately.