If ∫-0.15^(0.15)|100 x²-1| d x=k/3000, then k is equal to ___________.

Solution

Maths sir · 5 steps

Find where 100x² - 1 changes sign, split the integral there, and use evenness to halve the work.

1. Use evenness. The integrand is even and the limits are symmetric, so the integral equals 2∫₀^(0.15)|100x²-1|dx. Inside the modulus, 100x² - 1 = 0 at x = 0.1. It is negative on [0, 0.1] and positive on [0.1, 0.15].

   I = 2[∫₀^(0.1)(1-100x²)dx + ∫_(0.1)^(0.15)(100x²-1)dx]

2. First piece. [x - 100x³/3]₀^(0.1) = 0.1 - 1/30 = 1/15.
3. Second piece. [100x³/3 - x]_(0.1)^(0.15) = (0.1125 - 0.15) - (1/30 - 0.1) = -3/80 + 1/15 = 7/240.
4. Add and double. I = 2(16/240 + 7/240) = 23/120. Setting 23/120 = k/3000 gives k = (23 × 3000)/120.
5. So. k = 23 × 25 = 575.

The common slip: 225. Dropping the modulus and integrating 1 - 100x² straight across gives I = 0.075, so k = 225. The negative area cancels part of the positive area instead of adding to it.

The idea to keep. With a modulus inside an integral, first find the zeros of the expression and split the interval there. On every piece, take the expression with the sign that keeps it positive.

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