Area Between Curves — Question 6

Question 6 – Solution

1. Identify the upper curve on $[0,1]$.

Let $h(x)=e^x-x-1$. Then $h(0)=0$ and $$h'(x)=e^x-1\ge0\qquad(0\le x\le1).$$ Thus $e^x\ge x+1$, with equality only at $x=0$.

2. Use the given vertical boundaries.

$$A=\int_0^1\bigl(e^x-(x+1)\bigr)\,dx.$$

3. Integrate term by term.

$$\int e^x\,dx=e^x,\qquad \int x\,dx=\frac{x^2}{2},\qquad \int1\,dx=x.$$

4. Evaluate at $x=1$ and subtract the value at $x=0$.

$$\begin{align*} A&=\left[e^x-\frac{x^2}{2}-x\right]_0^1\\ &=\left(e-\frac12-1\right)-(1-0-0)\\ &=\boxed{e-\frac52}. \end{align*}$$

5. Check the sign and units.

The area is approximately $2.71828-2.5=0.21828$ square units. It is positive because $e^x$ lies above the line on $(0,1]$.