Let I_n = ∫₀¹ (1-x)ⁿeˣ dx. Show that, for all positive integers n, I_n = nI_(n-1) - 1. Find the exact value of I_4. By considering the area of the region enclosed by the x-axis, the y-axis and the curve with equation y = (1 – x)⁴eˣ in the interval 0 ≤ x ≤ 1, show that 1/24 < e < 11/4.

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2014). The mark or mark type is shown in brackets.

1. [M1A1] Using integration by parts (∫udv = uv - ∫vdu) with u = (1-x)ⁿ (du = -n(1-x)ⁿ⁻¹dx) and dv = eˣdx (v = eˣ): I_n = [(1-x)ⁿeˣ]₀¹ - ∫₀¹ eˣ(-n(1-x)ⁿ⁻¹dx).
2. [A1] I_n = (0 - 1ⁿe⁰) + n∫₀¹ (1-x)ⁿ⁻¹eˣdx = -1 + nI_(n-1). Thus, I_n = nI_(n-1) - 1. (AG)
3. [B1] First, calculate I₀ = ∫₀¹ eˣdx = [eˣ]₀¹ = e - 1.
4. [M1A1] Then, use the reduction formula: I₁ = 1I₀ - 1 = (e-1) - 1 = e-2. I₂ = 2I₁ - 1 = 2(e-2) - 1 = 2e-5. I₃ = 3I₂ - 1 = 3(2e-5) - 1 = 6e-16. I₄ = 4I₃ - 1 = 4(6e-16) - 1 = 24e-65.
5. [M1] The area of the region enclosed by y = (1-x)⁴eˣ, the x-axis, and the y-axis in [0,1] is I₄. Since y = (1-x)⁴eˣ is non-negative for x in [0,1], the area is positive. Thus, I₄ > 0. Also, (1-x)⁴eˣ < e for x in [0,1], so I₄ < ∫₀¹ e dx = e. More simply, observe that 0 < (1-x)⁴ < 1 and 1 < eˣ < e for x in (0,1). Therefore, 0 < (1-x)⁴eˣ < e. Integrating gives 0 < I₄ < e. The mark scheme uses a weaker upper bound for the area: the area of the rectangle 1x1 is 1. So 0 < I₄ < 1.
6. [A1] So, 0 < 24e - 65 < 1.
7. [A1] From 24e - 65 > 0, we get 24e > 65 ⇒ e > 65/24. From 24e - 65 < 1, we get 24e < 66 ⇒ e < 66/24 = 11/4. Combining these, 65/24 < e < 11/4. (AG)