Let f : R → R be defined as f(x) = e⁻ˣsinx. If F : [0, 1] → R is a differentiable function with that F(x) = ∫₀ˣ f(t)dt, then the value of ∫₀¹ (F'(x) + f(x))eˣdx lies in the interval

Solution

Maths sir · 5 steps

By the fundamental theorem F'(x) = f(x), so the eˣ cancels and you are left with a plain 2 sin x.

1. Use the theorem. Since F(x) = ∫₀ˣ f(t) dt with f continuous, the fundamental theorem gives F'(x) = f(x) = e⁻ˣ sin x.
2. Simplify the integrand. (F'(x) + f(x))eˣ = 2e⁻ˣ sin x · eˣ = 2 sin x. So the integral is ∫₀¹ 2 sin x dx = 2(1 - cos 1).
3. Bound cos 1. Use the series 1 - cos x = x²/2 - x⁴/24 + x⁶/720 - … at x = 1. The terms alternate in sign and keep shrinking, so stopping after a minus term gives a lower bound and stopping after a plus term gives an upper bound.

   1/2 - 1/24 ≤ 1 - cos 1 ≤ 1/2 - 1/24 + 1/720

4. Multiply by 2. The lower end becomes 1 - 1/12 = 330/360. The upper end becomes 330/360 + 1/360 = 331/360. As a check, 2(1 - cos 1) ≈ 0.9194.
5. So. The value lies in [330/360, 331/360], which is option (b).

The idea to keep. When an answer must lie in a narrow interval, use the alternating Taylor series of sin or cos. Stopping after a minus term gives one bound and stopping after a plus term gives the other.

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