Letf(x)=∫ x³√(3-x²)dx. If5f(√2)=-4, thenf(1) is equal to

Solution

Maths sir · 4 steps

Split x³ as x² · x: the lone x goes into dt and x² becomes 3 - t.

1. Choose the substitution. Put t = 3 - x², so dt = -2x dx and x dx = -dt/2. Write x³ dx = x² · x dx and replace x² by 3 - t.

   f(x) = -1/2∫ (3 - t)√t dt = -1/2(2t^(3/2) - 2/5t^(5/2)) + C = -t^(3/2) + 1/5t^(5/2) + C

2. Find C. At x = √2, t = 1, so f(√2) = -1 + 1/5 + C = -4/5 + C. The condition 5f(√2) = -4 gives f(√2) = -4/5, so C = 0.
3. Evaluate f(1). At x = 1, t = 2. Then t^(3/2) = 2√2 and t^(5/2) = 4√2.
4. So. f(1) = -2√2 + 4√2/5 = -10√2/5 + 4√2/5 = -6√2/5, which is option (a).

The idea to keep. When an odd power of x sits next to √(a - x²), keep one x for dt and turn the even power into an expression in t. Then use the given value to fix C.

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