That Is Close 4

Calculus Level 5

lim ⁡ t → 0 [ ∫ 0 1 ( b x + a ( 1 − x ) ) t d x ] 1 / t \large \lim_{t\to 0} \left[\int_0^{1} (bx + a(1-x))^{t} dx\right]^{1/t}

For 0 < a < b 0 < a < b , the limit above can be expressed as e − p 1 ( b b a − a ) p 2 b − a \large e^{-p_1} \left(b^{b}a^{-a}\right)^{\frac{p_2}{b-a}} where p 1 p_1 and p 2 p_2 are constants. Find p 1 + p 2 p_1 + p_2 .


The answer is 2.

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Ujjwal Mani Tripathi by Ujjwal Mani Tripathi , 21, India

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1 solution

Relevant wiki: L'Hopital's Rule - Problem Solving

I = ∫ 0 1 ( b x + a ( 1 − x ) ) t d x = ∫ 0 1 ( ( b − a ) x + a ) t d x Let u = ( b − a ) x + a , d u = ( b − a ) d x = ∫ a b u t b − a d u = b t + 1 − a t + 1 ( b − a ) ( t + 1 ) \begin{aligned} I & = \int_0^1 (bx+a(1-x))^t dx \\ & = \int_0^1 ((b-a)x+a)^t dx & \small {\color{#3D99F6}\text{Let }u = (b-a)x+a, \ du = (b-a) \ dx} \\ & = \int_a^b \frac{u^t}{b-a} du \\ & = \frac {b^{t+1}-a^{t+1}}{(b-a)(t+1)} \end{aligned}

L = lim ⁡ t → 0 [ ∫ 0 1 ( b x + a ( 1 − x ) ) t d x ] 1 / t = lim ⁡ t → 0 [ b t + 1 − a t + 1 ( b − a ) ( t + 1 ) ] 1 / t = lim ⁡ t → 0 exp ⁡ ( 1 t ln ⁡ [ b t + 1 − a t + 1 ( b − a ) ( t + 1 ) ] ) exp ⁡ ( x ) = e x = exp ⁡ ( lim ⁡ t → 1 ln ⁡ [ b t − a t ( b − a ) t ] t − 1 ) A 0/0 cases, L’H o ˆ pital’s rule applies. = exp ⁡ ( lim ⁡ t → 1 ( b − a ) t b t − a t ( ( b t ln ⁡ b − a t ln ⁡ a ) ( b − a ) t − ( b t − a t ) ( b − a ) ( ( b − a ) t ) 2 ) 1 ) Differentiate up and down. = exp ⁡ ( b ln ⁡ b − a ln ⁡ a b − a − 1 ) = exp ⁡ ( ln ⁡ b b b − a − ln ⁡ a a b − a − 1 ) = e − 1 ( b b a − a ) 1 b − a \begin{aligned} L & = \lim_{t \to 0} \left[\int_0^{1} (bx + a(1-x))^{t} dx\right]^{1/t} \\ & = \lim_{t \to 0} \left[\frac {b^{t+1}-a^{t+1}}{(b-a)(t+1)} \right]^{1/t} \\ & = \lim_{t \to 0} \exp \left( \frac 1t \ln \left[\frac {b^{t+1}-a^{t+1}}{(b-a)(t+1)} \right] \right) & \small {\color{#3D99F6} \exp(x) = e^x} \\ & = \exp \left( \lim_{t \to \color{#D61F06}1} \frac {\ln \left[\frac {b^{\color{#D61F06}t}-a^{\color{#D61F06}t}}{(b-a)\color{#D61F06}t} \right]}{\color{#D61F06}t-1} \right) & \small {\color{#3D99F6} \text{A 0/0 cases, L'Hôpital's rule applies.}} \\ & = \exp \left( \lim_{t \to 1} \frac {\frac {(b-a)t}{b^t-a^t} \left(\frac {(b^t\ln b - a^t \ln a)(b-a)t - (b^t-a^t)(b-a)}{((b-a)t)^2}\right)}1 \right) & \small {\color{#3D99F6} \text{Differentiate up and down.}} \\ & = \exp \left(\frac {b\ln b - a\ln a}{b-a} - 1 \right) \\ & = \exp \left(\ln b^{\frac b{b-a}} - \ln a^{\frac a{b-a}} - 1 \right) \\ & = e^{-1}\left(b^b a^{-a} \right)^{\frac 1{b-a}} \end{aligned}

⟹ p 1 + p 2 = 1 + 1 = 2 \implies p_1 + p_2 = 1+1 = \boxed{2}

Alternative Solution

L = lim ⁡ t → 0 [ b t + 1 − a t + 1 ( b − a ) ( t + 1 ) ] 1 / t = lim ⁡ t → 1 [ b t − a t ( b − a ) t ] 1 t − 1 The limit is of the form 1 ∞ = exp ⁡ ( lim ⁡ t → 1 1 t − 1 [ b t − a t ( b − a ) t − 1 ] ) A 0/0 cases, L’H o ˆ pital’s rule applies. = exp ⁡ ( lim ⁡ t → 1 1 1 [ ( b t ln ⁡ b − a t ln ⁡ a ) ( b − a ) t − ( b t − a t ) ( b − a ) ( ( b − a ) t ) 2 ] ) Differentiate up and down. = exp ⁡ ( b ln ⁡ b − a ln ⁡ a b − a − 1 ) = exp ⁡ ( ln ⁡ b b b − a − ln ⁡ a a b − a − 1 ) = e − 1 ( b b a − a ) 1 b − a \begin{aligned} L & = \lim_{t \to 0} \left[\frac {b^{t+1}-a^{t+1}}{(b-a)(t+1)} \right]^{1/t} \\ & = \lim_{t \to 1} \left[\frac {\color{#3D99F6}b^t-a^t}{\color{#D61F06}(b-a)t} \right]^{\color{#20A900}\frac 1{t-1}} & \small {\color{#3D99F6}\text{The limit is of the form }1^\infty} \\ & = \exp \left( \lim_{t \to 1} {\color{#20A900}\frac 1{t-1}} \left[\frac {\color{#3D99F6}b^t-a^t}{\color{#D61F06}(b-a)t} - 1 \right] \right) & \small {\color{#3D99F6} \text{A 0/0 cases, L'Hôpital's rule applies.}} \\ & = \exp \left( \lim_{t \to 1} \frac 11 \left[\frac {(b^t\ln b-a^t\ln a)(b-a)t-(b^t-a^t)(b-a)}{((b-a)t)^2} \right] \right) & \small {\color{#3D99F6} \text{Differentiate up and down.}} \\ & = \exp \left(\frac {b\ln b - a\ln a}{b-a} - 1 \right) \\ & = \exp \left(\ln b^{\frac b{b-a}} - \ln a^{\frac a{b-a}} - 1 \right) \\ & = e^{-1}\left(b^b a^{-a} \right)^{\frac 1{b-a}} \end{aligned}

⟹ p 1 + p 2 = 1 + 1 = 2 \implies p_1 + p_2 = 1+1 = \boxed{2}

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an amazing solution :)

Ujjwal Mani Tripathi - 4 years, 7 months ago

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