If `y=f(x)` makes positive intercepts of 2 and 1 unit on x and y-coordinates axes and encloses an area of `3/4` sq unit with the axes, then `int_0^2xf'(x)dx` equals

Options

A. `3/2`
B. `1`
C. `5/4`
D. `-3/4`

Answer

To solve the problem step-by-step, we need to find the value of the integral \( \int_0^2 x f'(x) \, dx \) given the conditions about the function \( f(x) \).

### Step 1: Understand the intercepts and area
The function \( f(x) \) intercepts the x-axis at \( x = 2 \) and the y-axis at \( y = 1 \). This means the points of intersection are \( (2, 0) \) and \( (0, 1) \). The area enclosed by the curve and the axes is given as \( \frac{3}{4} \) square units.

### Step 2: Set up the area under the curve
The area under the curve from \( x = 0 \) to \( x = 2 \) can be expressed as:
\[ \text{Area} = \int_0^2 f(x) \, dx = \frac{3}{4} \]

### Step 3: Use integration by parts
To find \( \int_0^2 x f'(x) \, dx \), we can use integration by parts. Let:
- \( u = x \) → \( du = dx \)
- \( dv = f'(x) \, dx \) → \( v = f(x) \)

Using integration by parts:
\[ \int u \, dv = uv - \int v \, du \]
we have:
\[ \int_0^2 x f'(x) \, dx = \left[ x f(x) \right]_0^2 - \int_0^2 f(x) \, dx \]

### Step 4: Evaluate the boundary term
Now we evaluate \( \left[ x f(x) \right]_0^2 \):
- At \( x = 2 \): \( f(2) = 0 \) (since it intersects the x-axis)
- At \( x = 0 \): \( f(0) = 1 \)

Thus,
\[ \left[ x f(x) \right]_0^2 = 2 \cdot f(2) - 0 \cdot f(0) = 2 \cdot 0 - 0 = 0 \]

### Step 5: Substitute back into the equation
Now substituting back, we get:
\[ \int_0^2 x f'(x) \, dx = 0 - \int_0^2 f(x) \, dx \]
Since we know \( \int_0^2 f(x) \, dx = \frac{3}{4} \):
\[ \int_0^2 x f'(x) \, dx = -\frac{3}{4} \]

### Final Answer
Thus, the value of \( \int_0^2 x f'(x) \, dx \) is:
\[ \boxed{-\frac{3}{4}} \]