Let (a, b) be the point of intersection of the curve x²=2 y and the straight line y-2 x-6=0 in the second quadrant. Then the integral I=∫ₐᵇ (9 x²)/(1+5ˣ) d x is equal to :

Solution

Maths sir · 3 steps

The limits come out as -2 and 2, so pair f(x) with f(-x) and the 5ˣ vanishes.

1. Find the limits. Put y = 2x + 6 into x² = 2y to get x² - 4x - 12 = 0, so x = 6 or x = -2. The second quadrant needs x < 0, so x = -2 and y = 2. That gives a = -2 and b = 2.
2. Use the symmetric limits. On [-a, a] you can replace x by -x without changing the integral. Then 1/(1+5⁻ˣ) = 5ˣ/(1+5ˣ), so I = ∫₋₂² (9x² · 5ˣ)/(1+5ˣ) dx. Adding this to the original form of I cancels the denominator.

   2I = ∫₋₂² 9x² · (1 + 5ˣ)/(1 + 5ˣ) dx = ∫₋₂² 9x² dx = [3x³]₋₂² = 24 + 24 = 48

3. So. I = 48/2 = 24, which is option (c).

The idea to keep. When the limits are -a to a and you see 1 + kˣ in the denominator, write the integral again with x → -x and add. The kˣ cancels and only the even part is left.

Answer from the NTA final answer key. Solution by PrepWiser.