Let (a, b) be the point of intersection of th | vedclass

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(A) Given the curve $x^2=2y$ and the line $y=2x+6$. Substituting $y$ from the line equation into the curve equation: $x^2=2(2x+6) \Rightarrow x^2=4x+1

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$24$

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(A) Given the curve $x^2=2y$ and the line $y=2x+6$. Substituting $y$ from the line equation into the curve equation: $x^2=2(2x+6) \Rightarrow x^2=4x+12$ $x^2-4x-12=0 \Rightarrow (x-6)(x+2)=0$ So,$x=6$ or $x=-2$. For $x=6$,$y=2(6)+6=18$. For $x=-2$,$y=2(-2)+6=2$. The points of intersection are $(6, 18)$ and $(-2, 2)$. Since the point $(a, b)$ lies in the second quadrant,we have $a=-2$ and $b=2$. We need to evaluate $I=\int_{-2}^2 \frac{9x^2}{1+5^x} dx$. Using the property $\int_{-k}^k f(x) dx = \int_0^k (f(x)+f(-x)) dx$: $I = \int_{-2}^2 \frac{9x^2}{1+5^x} dx$. Note that $\frac{9x^2}{1+5^x} + \frac{9(-x)^2}{1+5^{-x}} = \frac{9x^2}{1+5^x} + \frac{9x^2 \cdot 5^x}{5^x+1} = \frac{9x^2(1+5^x)}{1+5^x} = 9x^2$. Thus,$I = \int_0^2 9x^2 dx = [3x^3]_0^2 = 3(8) - 3(0) = 24$.

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:

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