Find tangent of $2y=\ln{(x+1)}$ and calculate area between tangent,function and abscissa

Question

We are given a curve $2y=\ln{(x+1)}$

1. Find equation of tangent of given curve parallel to line $4y-2e^{-2}x-3=0$
2. Calculate area between curve,tangent and abscissa.
My solution:

1. Curve: $y=\frac{\ln{(x+1)}}{2}$

We have to find a tangent with slope $k=\frac{1}{2} e^{-2}$

$y'=\frac{1}{2} \frac{1}{x+1}$

We have $\frac{1}{2} \frac{1}{x+1}=\frac{1}{2} e^{-2}$, so $x=e^2-1$. For that $x$, $y=1$ and $y'=\frac{1}{2} e^{-2}$. Tangent has formula $g(x)=\frac{1}{2} e^{-2}(x-e^2+1)+1.$

2. I drew a picture, and i don't know how to calculate the area. I know there must be some combination of integrals, i was hoping if someone can draw a picture and give me solution. Thank you.

EDIT: Maybe I have wrong picture, i tried not to use WolframAlpha because we can't use it on exams. How do you even draw something like this?

Answer 1

You only have to know two things:

• The graph of $y = \ln (x)$. Graph of $y = \ln(x+a)$ is just the graph of $\ln(x)$ moved $a$ units to the left.
• What does it means to be tangent.

I suppose the area they want is the shaded one. It is $$ \int_0^{e^2-1} \left(\left[e^{-2}/2 (x - e^2 + 1) + 1\right] - \left[\frac12 \ln(x+1) \right]\right) \mathrm dx$$

Answer 2

Tangent formula you gave appears to be correct (unless I've forgotten GCSE!).

Assuming the area is the same as JnxF's prediction, integral evaluates to $\frac{e^4-4e^2-1}{4e^2}$