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JEE Main 20252 AprilEvening Shift
Exam
JEE Main
Year
2025
Date
2 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Definite Integration

JEE Main 2025 · 2 April · Evening Shift · Mathematics · Definite Integration

Let (a, b) be the point of intersection of the curve x^2=2 y and...

Let (a,b)(a, b) be the point of intersection of the curve x2=2yx^2=2 y and the straight line y−2x−6=0y-2 x-6=0 in the second quadrant. Then the integral I=∫ab9x21+5xdxI=\int_a^b \frac{9 x^2}{1+5^x} d x is equal to :

Options

  1. A2424
  2. B2727
  3. C1818
  4. D2121

Correct answer

A. 2424

Step-by-step solution

x2=2yy=2x+6x2=4x+12\begin{aligned} & x^2=2 y & y=2 x+6 \\ & x^2=4 x+12\end{aligned}<br/>x2−4x−12=0⇒x=6y=18∣  if x=−2y=2<br/>\left.x^2-4 x-12=0 \Rightarrow \begin{gathered}x=6 \\ y=18\end{gathered} \right\rvert\, \begin{gathered}\text{ if } x=-2 \\ y=2\end{gathered}
∴(6,18)&(−2,2)\therefore(6,18) \&(-2,2)
Here (6,18)(6,18) Rejected because (a,b)(a, b) lies in 2nd 2^{\text{nd }} quadrant
∴a=−2 b=2∴I=∫−229x21+5xdx=∫−229⋅5x⋅x21+5xdx∴2I=∫−229x2dx=18∫02x2dx=18(x33)022I=48∴I=24\begin{aligned} & \therefore \mathrm{a}=-2 & \mathrm{~b}=2 \\ & \therefore \mathrm{I}=\int_{-2}^2 \frac{9 \mathrm{x}^2}{1+5^{\mathrm{x}}} \mathrm{dx}=\int_{-2}^2 \frac{9 \cdot 5^{\mathrm{x}} \cdot \mathrm{x}^2}{1+5^{\mathrm{x}}} \mathrm{dx} \\ & \therefore 2 \mathrm{I}=\int_{-2}^2 9 \mathrm{x}^2 \mathrm{dx}=18 \int_0^2 \mathrm{x}^2 \mathrm{dx}=18\left(\frac{\mathrm{x}^3}{3}\right)_0^2 \\ & 2 \mathrm{I}=48 \\ & \therefore \mathrm{I}=24\end{aligned}
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