Is this just another way of asking for the definite integral

Question

Is this the same question just expressed differently?

3. Find

(a) $\displaystyle \int_2^4 \left(\frac{2}{x-1}-1\right)dx$

(b) The area between the curve $y = \left(\dfrac{2}{x-1}-1\right)dx$ and the $x$-axis over the interval $[2,4]$

Answer 1

The second one is really asking for:

$$\int_2^4 \ \left| \ \frac{2}{x-1} - 1 \ \right| \ \text{d}x$$

This is because the original integral subtracts area while the function is below the horizontal axis—the reason why, for instance:

$$\int_{-2\pi}^{2\pi} \sin(x) \ \text{d}x = 0$$

Answer 2

No. The function $\frac{2}{x-1}-1$ changes sign at $x = 3$. The integral, version (a), will partially cancel the positive (signed) area with the negative (signed) area. The area between $\frac{2}{x-1}-1$ and the $y$-axis, version (b), does not have a negative contribution for $x \in [3,4]$.