Finding the area under $\frac{3}{2x+1}$ and $3x-2$

Question

Find the area between these curves.

$$y=\dfrac{3}{2x+1},\qquad y=3x-2;\qquad x=2\quad \text{et} \quad y=0 $$

Indeed,

I calculate the integral of the blue function between $1$ and $2$. Then, I will calculate the area of the triangle between the yellow line and the x-axis on the interval $\dfrac{2}{3}$ and $1$, which gives me:

$$\int _{\frac{3}{2}}^1\left(3x-2\right)dx+\int _1^2\frac{3}{2x+1}dx=\frac{3}{2}\ln \left(\frac{5}{3}\right)-\frac{7}{8}$$

Am I right in the way how to find calculate the area and if that so that $\frac{3}{2}\ln \left(\frac{5}{3}\right)-\frac{7}{8}$ is it correct?

Answer 1

i think the first integral must be $$\int_{2/3}^{1}(3x-2)dx$$

Answer 2

As mentioned in the comments, you are almost correct.

But your first integral should start at $2/3$, not $3/2$. So you should have $$\int _{\frac{2}{3}}^1\left(3x-2\right)dx+\int _1^2 \frac{3}{2x+1} dx = \frac{3}{2}\ln \left(\frac{5}{3}\right) + \frac{1}{6}.$$

As a sanity check, this makes much more sense for a few reasons. Firstly, $3x-2$ is positive in your picture and so it doesn't make sense for its integral contribution to be negative. Secondly, $3/2$ is contained in the interval $[1,2]$, which is where the other integral is accounting for the area already.