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Step 1: Solve the given differential equation
The differential equation is
\frac{dy}{dx} = 2(x + 1)\,.
Integrate both sides with respect to x :
y = \int 2(x + 1) \, dx = \int (2x + 2) \, dx = x^2 + 2x + C,
where C is the constant of integration.
Step 2: Interpret the area bounded by the parabola and the x-axis
According to the problem, the area bounded by y = x^2 + 2x + C and the x-axis is
\frac{4\sqrt{8}}{3} \,.
The x-intercepts of the parabola come from solving
x^2 + 2x + C = 0.
Let the roots of this quadratic be \alpha and \beta . Then the discriminant D is given by
D = b^2 - 4ac = (2)^2 - 4(1)(C) = 4 - 4C.
Step 3: Use the geometry of the bounded area
From the geometric approach, the parabola with roots \alpha and \beta encloses some region above or below the x-axis. The figure (provided in the problem) shows a rectangle that helps us find the area of the bounded region. An illustrative image is:
• |\alpha - \beta| = \sqrt{D} (the distance between the roots).
• The “height” corresponding to that rectangle is related to -\frac{D}{4a} (in this case a=1 ).
Hence, the area of the rectangle formed is
\text{Area}_\text{rectangle} = (\sqrt{D}) \times \left(\frac{D}{4}\right) = \frac{D\sqrt{D}}{4}.
For a standard property of parabolas, the region actually bounded by the parabola and the x-axis is
\frac{2}{3} \times \text{Area}_\text{rectangle}.
Therefore,
\frac{4\sqrt{8}}{3} = \frac{2}{3} \times \frac{D\sqrt{D}}{4}.
Step 4: Solve for the discriminant D
Simplify the equation:
\frac{4\sqrt{8}}{3} = \frac{2}{3} \times \frac{D\sqrt{D}}{4}
\quad\Longrightarrow\quad
4\sqrt{8} = 2 \times \frac{D\sqrt{D}}{4}
\quad\Longrightarrow\quad
4\sqrt{8} = \frac{D\sqrt{D}}{2}.
Multiply both sides by 2:
8\sqrt{8} = D\sqrt{D}.
To satisfy D\sqrt{D} = 8\sqrt{8} , a suitable solution is
D = 8.
Step 5: Find the constant C
Recall
D = 4 - 4C = 8.
Hence
4 - 4C = 8
\quad\Longrightarrow\quad
-4C = 4
\quad\Longrightarrow\quad
C = -1.
Step 6: Final function and calculate y(1)
With C = -1, the function is
y(x) = x^2 + 2x - 1.
Thus,
y(1) = 1^2 + 2\cdot 1 -1 = 1 + 2 - 1 = 2.
Therefore, the value of y(1) is
2.
