It's All Equal Areas

Calculus Level 4

Let e e be Euler's number and x > 1 x > -1 .

Let f ( x ) = lim n j = 1 n ( j n ) n ( 1 e x ) n j f(x) = \lim_{n \rightarrow \infty} \sum_{j = 1}^{n} (\dfrac{j}{n})^n (\dfrac{1}{e^x})^{n - j} and g ( x ) = a x 2 + b x + c g(x) = ax^2 + bx + c .

If f ( 0 ) = g ( 0 ) f(0) = g(0) , f ( 1 ) = g ( 1 ) f(1) = g(1) and 0 1 f ( x ) = 0 1 g ( x ) \int_{0}^{1} f(x) = \int_{0}^{1} g(x) , find the area bounded by f ( x ) f(x) and g ( x ) g(x) on [ 1 , 2 ] [1,2] to eight decimal places.


The answer is 0.12937127.

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1 solution

Rocco Dalto
Jun 30, 2018

f ( x ) = lim n j = 1 n ( j n ) n ( 1 e x ) n j = f(x) = \lim_{n \rightarrow \infty} \sum_{j = 1}^{n} (\dfrac{j}{n})^n (\dfrac{1}{e^x})^{n - j} = lim n j = 0 n 1 ( 1 j n ) n ( 1 e x ) j = \lim_{n \rightarrow \infty} \sum_{j = 0}^{n - 1} (1 - \dfrac{j}{n})^n (\dfrac{1}{e^x})^{j} = j = 0 ( 1 e x + 1 ) j = e x + 1 e x + 1 1 \sum_{j = 0}^{\infty} (\dfrac{1}{e^{x + 1}})^j = \dfrac{e^{x + 1}}{e^{x + 1} - 1} on x > 1 x > -1 .

Let g ( x ) = a x 2 + b x + c g(x) = ax^2 + bx + c .

g ( 0 ) = f ( 0 ) c = e e 1 g(0) = f(0) \implies c = \dfrac{e}{e - 1} and f ( 1 ) = g ( 1 ) a + b = e e 2 1 f(1) = g(1) \implies \boxed{a + b = -\dfrac{e}{e^2 - 1}}

0 1 g ( x ) = 0 1 f ( x ) = ln ( e x + 1 1 ) 0 1 = ln ( e + 1 ) ln ( e + 1 ) = 0 1 a x 2 + b x + e e 1 d x = a 3 + b 2 + e e 1 \int_{0}^{1} g(x) = \int_{0}^{1} f(x) = \ln(e^{x + 1} - 1)|_{0}^{1} = \ln(e + 1) \implies \ln(e + 1) = \int_{0}^{1} ax^2 + bx + \dfrac{e}{e - 1} dx = \dfrac{a}{3} + \dfrac{b}{2} + \dfrac{e}{e - 1} \implies

2 a + 3 b = 6 ln ( e + 1 ) 6 e e 1 \boxed{2a + 3b = 6\ln(e + 1) - \dfrac{6e}{e - 1}}

a + b = e e 2 1 \boxed{a + b = -\dfrac{e}{e^2 - 1}}

Solving the system above we obtain:

a = 6 e 2 + 3 e 6 ( e 2 1 ) ln ( e + 1 ) e 2 1 a = \dfrac{6e^2 + 3e - 6(e^2 - 1)\ln(e + 1)}{e^2 - 1} and b = 6 ( e 2 1 ) ln ( e + 1 ) 6 e 2 4 e e 2 1 b = \dfrac{6(e^2 - 1)\ln(e + 1) - 6e^2 - 4e}{e^2 - 1} .

For 1 2 g ( x ) f ( x ) d x \int_{1}^{2} g(x) - f(x) dx :

1 2 f ( x ) d x = ln ( e x + 1 1 ) 1 2 = ln ( e 2 + e + 1 e + 1 ) \int_{1}^{2} f(x) dx = \ln(e^{x + 1} - 1)|_{1}^{2} = \ln(\dfrac{e^2 + e + 1}{e + 1})

and

1 2 g ( x ) d x = 7 a 3 + 3 b 2 + c = 14 a + 9 b + 6 c 6 = \int_{1}^{2} g(x) dx = \dfrac{7a}{3} + \dfrac{3b}{2} + c = \dfrac{14a + 9b + 6c}{6} = 36 e 2 + 12 e 30 ( e 2 1 ) ln ( e + 1 ) 6 ( e 2 1 ) \dfrac{36e^2 + 12e - 30(e^2 - 1)\ln(e + 1)}{6(e^2 - 1)}

1 2 g ( x ) f ( x ) d x = 36 e 2 + 12 e 30 ( e 2 1 ) ln ( e + 1 ) 6 ( e 2 1 ) ln ( e 2 + e + 1 e + 1 ) 0.12937127 \implies \int_{1}^{2} g(x) - f(x) dx = \dfrac{36e^2 + 12e - 30(e^2 - 1)\ln(e + 1)}{6(e^2 - 1)} - \ln(\dfrac{e^2 + e + 1}{e + 1}) \approx \boxed{0.12937127} .

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