Area Between Curves — Question 9

Question 9 – Solution

1. Describe the intersections.

Let $h(x)=\ln(1+x)-x/2$, with domain $x>-1$. Then $$h'(x)=\frac{1-x}{2(1+x)}.$$ So $h$ increases up to $x=1$ and decreases thereafter. Since $h(0)=0$, $h(1)>0$, and $h(x)\to-\infty$ as $x\to\infty$, the only roots are $0$ and a unique number $a>1$ satisfying $$\ln(1+a)=\frac a2,\qquad a\approx2.512862417.$$ The nonzero root has no elementary closed form; this equation defines $a$ exactly.

2. Set up the area.

On $(0,a)$, $h(x)>0$, so the logarithm is above the line. $$A=\int_0^a\left(\ln(1+x)-\frac x2\right)dx.$$

3. Find an antiderivative.

With $u=1+x$, integration by parts gives $$\int\ln(1+x)\,dx=(1+x)\ln(1+x)-x+C.$$ Also, $\int(x/2)\,dx=x^2/4$. At $x=0$, all terms below are zero.

4. Evaluate, then substitute $\ln(1+a)=a/2$.

$$\begin{align*} A&=\left[(1+x)\ln(1+x)-x-\frac{x^2}{4}\right]_0^a\\ &=(1+a)\frac a2-a-\frac{a^2}{4}\\ &=\boxed{\frac{a(a-2)}4}\approx0.322188173. \end{align*}$$

5. Interpret the exact answer.

The boxed expression is exact with $a$ defined above; using its numerical approximation gives $0.322188173$ square units.