What is the area of the region enclosed by the graphs of $y=e^x-2$, $y=\sin x$, and $x=0$ ?

Key concept: To find the enclosed area, identify the intersection point of the two curves and integrate the top function minus the bottom function.

Step by step

We want the area enclosed by

$y=e^x-2, \quad y=\sin x, \quad x=0.$

Step 1: Find the intersection point

The curves meet where

$e^x-2=\sin x.$

This equation has a solution between $0$ and $1$. Let that intersection be at $x=a$.

Step 2: Determine which curve is on top

For $x$ between $0$ and $a$, the sine curve is above the exponential curve, so the area is

$A=\int_0^a \left(\sin x-(e^x-2)\right)dx.$

That is

$A=\int_0^a (\sin x-e^x+2)\,dx.$

Step 3: Integrate

$A=\left[-\cos x-e^x+2x\right]_0^a.$

So

= -\cos a-e^a+2a+2.$$ Using the intersection value numerically gives $$A\approx 0.745.$$ Therefore, the area is $$\boxed{0.745}$$ which is choice **C**. ### Pitfall alert A common mistake is to integrate from $0$ to a guessed endpoint without first locating the intersection of the two curves. Another error is reversing top and bottom, which would make the area negative before taking the absolute value. ### Try different conditions If the vertical line $x=0$ were replaced by another boundary such as $x=c$, then the limits of integration would change accordingly. The same method still applies: find where the curves intersect and subtract bottom from top over the bounded interval. ### Further reading area between curves, intersection point, definite integral