The curve `u=a sqrtx+bx` passes through the point (1, 2) and the area enclosed by the curve, the
x - axis and the line x = 4 is 8 sq. units. Then

Options

A. `a+2b=1`
B. `2a+b=1`
C. `a=-1, b=3`
D. `a=3, b=-1`

Answer

To solve the problem step by step, we need to find the values of \( a \) and \( b \) for the curve \( u = a \sqrt{x} + bx \) given the conditions that it passes through the point (1, 2) and that the area enclosed by the curve, the x-axis, and the line \( x = 4 \) is 8 square units.

### Step 1: Use the point (1, 2) to create an equation.
Since the curve passes through the point (1, 2), we can substitute \( x = 1 \) and \( u = 2 \) into the equation:

\[ 2 = a \sqrt{1} + b(1) \]

This simplifies to:

\[ 2 = a + b \quad \text{(Equation 1)} \]

### Step 2: Set up the area under the curve.
The area \( A \) under the curve from \( x = 0 \) to \( x = 4 \) can be expressed as:

\[ A = \int_0^4 (a \sqrt{x} + bx) \, dx \]

Given that this area equals 8 square units, we have:

\[ \int_0^4 (a \sqrt{x} + bx) \, dx = 8 \]

### Step 3: Calculate the integral.
We can split the integral:

\[ A = \int_0^4 a \sqrt{x} \, dx + \int_0^4 bx \, dx \]

Calculating each part:

1. For \( \int_0^4 a \sqrt{x} \, dx \):
\[ \int \sqrt{x} \, dx = \frac{2}{3} x^{3/2} \]
Evaluating from 0 to 4:
\[ = a \left[ \frac{2}{3} (4)^{3/2} - 0 \right] = a \left[ \frac{2}{3} \cdot 8 \right] = \frac{16a}{3} \]

2. For \( \int_0^4 bx \, dx \):
\[ \int x \, dx = \frac{x^2}{2} \]
Evaluating from 0 to 4:
\[ = b \left[ \frac{(4)^2}{2} - 0 \right] = b \cdot 8 = 8b \]

Combining these results, we have:

\[ A = \frac{16a}{3} + 8b = 8 \]

### Step 4: Set up the second equation.
From the area equation, we can rearrange it:

\[ \frac{16a}{3} + 8b = 8 \]

Multiplying through by 3 to eliminate the fraction:

\[ 16a + 24b = 24 \quad \text{(Equation 2)} \]

### Step 5: Solve the system of equations.
Now we have two equations:

1. \( a + b = 2 \) (Equation 1)
2. \( 16a + 24b = 24 \) (Equation 2)

From Equation 1, we can express \( b \) in terms of \( a \):

\[ b = 2 - a \]

Substituting this into Equation 2:

\[ 16a + 24(2 - a) = 24 \]

Expanding and simplifying:

\[ 16a + 48 - 24a = 24 \]
\[ -8a + 48 = 24 \]
\[ -8a = 24 - 48 \]
\[ -8a = -24 \]
\[ a = 3 \]

Now substituting \( a = 3 \) back into Equation 1 to find \( b \):

\[ 3 + b = 2 \implies b = 2 - 3 = -1 \]

### Final Result:
Thus, we have:

\[ a = 3, \quad b = -1 \]