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Question

Let $(a, b)$ be the point of intersection of the curve $x^2 = 2y$ and the straight line $y -2x-6=0$ in the second quadrant. Then the integral $I = \int_{a}^{b} \frac{9x^2}{1 + 5^x} dx$ is equal to :

The correct answer is
24
Solution Explanation
Step 1: Find the point of intersection in the second quadrant.
The curve is \[ x^2 = 2y \]
The line is \[ y - 2x - 6 = 0 \] which can be written as \[ y = 2x + 6 \]
Substitute the value of \(y\) from the line into the curve: \[ x^2 = 2(2x+6) \] \[ x^2 = 4x + 12 \] \[ x^2 - 4x - 12 = 0 \]
Factorizing: \[ (x-6)(x+2)=0 \]
So, the values of \(x\) are \[ x=6 \quad \text{or} \quad x=-2 \]
Now find the corresponding values of \(y\) using \[ y=2x+6 \]
For \(x=6\), \[ y=2(6)+6=18 \] So one point is \((6,18)\), which lies in the first quadrant.
For \(x=-2\), \[ y=2(-2)+6=2 \] So the other point is \((-2,2)\), which lies in the second quadrant.
Hence, in the second quadrant, \[ (a,b)=(-2,2) \] Therefore, \[ a=-2,\quad b=2 \]
Step 2: Write the integral.
Thus, \[ I=\int_{a}^{b}\frac{9x^2}{1+5^x}\,dx=\int_{-2}^{2}\frac{9x^2}{1+5^x}\,dx \]
Step 3: Use the property \(x \to -x\).
Let \[ I=\int_{-2}^{2}\frac{9x^2}{1+5^x}\,dx \]
Replacing \(x\) by \(-x\), we get \[ I=\int_{-2}^{2}\frac{9x^2}{1+5^{-x}}\,dx \]
Now, \[ \frac{1}{1+5^{-x}}=\frac{5^x}{1+5^x} \] So, \[ I=\int_{-2}^{2}\frac{9x^2\cdot 5^x}{1+5^x}\,dx \]
Step 4: Add the two expressions for \(I\).
Now add \[ I=\int_{-2}^{2}\frac{9x^2}{1+5^x}\,dx \] and \[ I=\int_{-2}^{2}\frac{9x^2\cdot 5^x}{1+5^x}\,dx \]
Then, \[ 2I=\int_{-2}^{2}\left(\frac{9x^2}{1+5^x}+\frac{9x^2\cdot 5^x}{1+5^x}\right)dx \]
Taking \(9x^2\) common: \[ 2I=\int_{-2}^{2}9x^2\left(\frac{1+5^x}{1+5^x}\right)dx \] \[ 2I=\int_{-2}^{2}9x^2\,dx \]
Step 5: Evaluate the integral.
\[ 2I=9\int_{-2}^{2}x^2\,dx \] \[ 2I=9\left[\frac{x^3}{3}\right]_{-2}^{2} \] \[ 2I=3\left(2^3-(-2)^3\right) \] \[ 2I=3(8+8)=48 \]
Therefore, \[ I=24 \]
Final Answer:
\[ \boxed{24} \]
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