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Step-by-Step Solution
Step 1: Understand the Given Information
We have a function $y = f(x)$ that:
Crosses the $x$-axis at $x = 2$, so $f(2) = 0$.
Encloses an area of $\tfrac{3}{4}$ square units with the coordinate axes over the interval $x = 0$ to $x = 2$, meaning $\displaystyle \int_{0}^{2} f(x)\,dx = \tfrac{3}{4}$.
Step 2: Set Up the Required Integral
We need to find $\displaystyle \int_{0}^{2} x\,f'(x)\,dx$.
Step 3: Apply Integration by Parts (Conceptual)
Recall the integration by parts formula for $\int u\,dv = uv - \int v\,du$. Here, we can think of:
\[
u = x
\quad\text{and}\quad
dv = f'(x)\,dx.
\]
Then:
\[
du = dx
\quad\text{and}\quad
v = f(x).
\]
Thus,
\[
\int_{0}^{2} x\,f'(x)\,dx
= \bigl[\,x\,f(x)\bigr]_{0}^{2} \;-\;\int_{0}^{2} f(x)\,dx.
\]
Step 4: Evaluate Each Part
Term $\bigl[x\,f(x)\bigr]_{0}^{2}$:
\[
= \bigl(2\,f(2)\bigr) - \bigl(0\,f(0)\bigr) = 2 \cdot 0 - 0 = 0
\quad\text{(since } f(2) = 0\text{).}
\]
Term $-\int_{0}^{2} f(x)\,dx$:
\[
= - \left(\tfrac{3}{4}\right).
\]
Adding them gives:
\[
\int_{0}^{2} x\,f'(x)\,dx = 0 - \tfrac{3}{4} = -\tfrac{3}{4}.
\]
Step 5: Conclusion
Therefore,
\[
\int_{0}^{2} x\,f'(x)\,dx
= -\tfrac{3}{4}.
\]
Hence, the correct answer is $\displaystyle -\tfrac{3}{4}$.
