Area with Parametric Equations — Question 6

Question 6 – Solution

Solution

1. On $0\le t\le\ln4$, $x=e^t$ increases from $1$ to $4$ because $$x'(t)=e^t>0.$$ Therefore, the requested area under the curve is directly $\int y\,dx$.

2. Substitute the parametric expressions: $$A=\int_0^{\ln4}y(t)x'(t)\,dt =\int_0^{\ln4}e^{-t}e^t\,dt.$$

3. The exponential factors cancel: $$e^{-t}e^t=1.$$ Hence $$A=\int_0^{\ln4}1\,dt =\left[t\right]_0^{\ln4} =\boxed{\ln4}.$$

4. As a check, eliminating $t$ gives $y=1/x$, and $\int_1^4(1/x)\,dx=\ln4$, the same result.