Area Bonanza 3

Calculus Level 4

Let ∣ x ∣ < 1 |x| < 1 .

If f ( x ) = ∑ n = 1 ∞ n 4 x n f(x) = \sum_{n = 1}^{\infty} n^4 x^n and g ( x ) = ∑ n = 1 ∞ n 3 x n g(x) = \sum_{n = 1}^{\infty} n^3 x^n , find the area of the region bounded by f f and g g on [ − 7 + 33 2 , 0 ] [\dfrac{-7 + \sqrt{33}}{2},0] .


The answer is 0.03635094.

Rocco Dalto by Rocco Dalto , 67, USA

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1 solution

Rocco Dalto
May 8, 2018

Let ∣ x ∣ < 1 |x| < 1 .

∑ n = 1 ∞ n x n = ∑ j = 1 ∞ ∑ n = j ∞ x n = 1 1 − x ∑ j = 1 ∞ x j = 1 1 − x ( x ) 1 1 − x = x ( 1 − x ) 2 \sum_{n = 1}^{\infty} n x^n = \sum_{j = 1}^{\infty} \sum_{n = j}^{\infty} x^n = \dfrac{1}{1 - x}\sum_{j = 1}^{\infty} x^{j} = \dfrac{1}{1 - x}(x)\dfrac{1}{1 - x} = \dfrac{x}{(1 - x)^2}

( x x − 1 ) ( 1 x ) ( x x − 1 ) = x ( x − 1 ) 2 . (\dfrac{x}{x - 1})(\dfrac{1}{x})(\dfrac{x}{x - 1}) = \dfrac{x}{(x - 1)^2}.

∑ n = 1 ∞ n 2 x n = ∑ j = 1 ∞ ( ∑ n = j ∞ n x n ) = \sum_{n = 1}^{\infty} n^2 x^n = \sum_{j = 1}^{\infty} (\sum_{n = j}^{\infty} n x^n) = 1 1 − x ∑ j = 1 ∞ ( j x j ) + x ( 1 − x ) 2 ∑ j = 1 ∞ x j = \dfrac{1}{1 - x}\sum_{j = 1}^{\infty} (j x^j) + \dfrac{x}{(1 - x)^2}\sum_{j = 1}^{\infty} x^j =

( 1 1 − x ) ( x ( 1 − x ) 2 ) + ( x ( 1 − x ) 2 ) ( x 1 − x ) = (\dfrac{1}{1 - x})(\dfrac{x}{(1 - x)^2}) + (\dfrac{x}{(1 - x)^2})(\dfrac{x}{1 - x}) = x 2 + x ( 1 − x ) 3 \dfrac{x^2 + x}{(1 - x)^3} .

g ( x ) = ∑ n = 1 ∞ n 3 x n = ∑ j = 1 ∞ ∑ n = j ∞ n 2 x n = g(x) = \sum_{n = 1}^{\infty} n^3 x^n = \sum_{j = 1}^{\infty} \sum_{n = j}^{\infty} n^2 x^{n} = 1 1 − x ∑ j = 1 ∞ j 2 x j + 2 x ( 1 − x ) 2 ∑ j = 1 ∞ j x j + x 2 + x ( 1 − x ) 3 ∑ j = 1 ∞ x j = ( 1 1 − x ) ( x 2 + x ( 1 − x ) 3 ) + \dfrac{1}{1 - x}\sum_{j = 1}^{\infty} j^2 x^j + \dfrac{2x}{(1 - x)^2}\sum_{j = 1}^{\infty} j x^j + \dfrac{x^2 + x}{(1 - x)^3}\sum_{j = 1}^{\infty} x^j = (\dfrac{1}{1 - x})(\dfrac{x^2 + x}{(1 - x)^3}) + ( 2 x ( 1 − x ) 2 ) ( x ( 1 − x ) 2 ) + ( x 2 + x ( 1 − x ) 3 ) ( x 1 − x ) = (\dfrac{2x}{(1 - x)^2})(\dfrac{x}{(1 - x)^2}) + (\dfrac{x^2 + x}{(1 - x)^3})(\dfrac{x}{1 - x}) = x 3 + 4 x 2 + x ( 1 − x ) 4 \dfrac{x^3 + 4x^2 + x}{(1 - x)^4} .

f ( x ) = ∑ n = 1 ∞ n 4 x n = ∑ j = 1 ∞ ∑ n = j ∞ n 3 x n = f(x) = \sum_{n = 1}^{\infty} n^4 x^n = \sum_{j = 1}^{\infty} \sum_{n = j}^{\infty} n^3 x^{n} = 1 1 − x ∑ j = 1 ∞ j 3 x j + 3 x ( 1 − x ) 2 ∑ j = 1 ∞ j 2 x j + 3 ( x 2 + x ) ( 1 − x ) 3 ∑ j = 1 ∞ j x j + x 3 + 4 x 2 + x ( 1 − x ) 4 ∑ j = 1 ∞ x j = \dfrac{1}{1 - x}\sum_{j = 1}^{\infty} j^3 x^j + \dfrac{3x}{(1 - x)^2}\sum_{j = 1}^{\infty} j^2 x^j + \dfrac{3(x^2 + x)}{(1 - x)^3}\sum_{j = 1}^{\infty} j x^j + \dfrac{x^3 + 4x^2 + x}{(1 - x)^4}\sum_{j = 1}^{\infty} x^j =

= ( 1 1 − x ) ( x 3 + 4 x 2 + x ( 1 − x ) 4 ) + = (\dfrac{1}{1 - x})( \dfrac{x^3 + 4x^2 + x}{(1 - x)^4}) + ( 3 x ( 1 − x ) 2 ) ( x 2 + x ( 1 − x ) 3 ) + ( 3 ( x 2 + x ) ( 1 − x ) 3 ) ( x ( 1 − x ) 2 ) + (\dfrac{3x}{(1 - x)^2})(\dfrac{x^2 + x}{(1 - x)^3}) + (\dfrac{3(x^2 + x)}{(1 - x)^3})(\dfrac{x}{(1 - x)^2}) + x 3 + 4 x 2 + x ( 1 − x ) 4 ( x 1 − x ) = x 4 + 11 x 3 + 11 x 2 + x ( 1 − x ) 5 \dfrac{x^3 + 4x^2 + x}{(1 - x)^4}(\dfrac{x}{1 - x}) = \dfrac{x^4 + 11x^3 + 11x^2 + x}{(1 - x)^5} .

Let u = 1 − x ⟹ d u = − d x ⟹ ∫ − 7 + 33 2 0 g ( x ) d x = − ∫ 1 9 − 33 2 u 3 − 7 u 2 + 12 u − 6 u 4 d u = u = 1 -x \implies du = -dx \implies \int_{\frac{-7 + \sqrt{33}}{2}}^{0} g(x) dx = -\int_{1}^{\frac{9 - \sqrt{33}}{2}} \dfrac{u^3 - 7u^2 + 12u - 6}{u^4} du = − ∫ 1 9 − 33 2 1 u − 7 u − 2 + 12 u − 3 − 6 u − 4 d u = -\int_{1}^{\frac{9 - \sqrt{33}}{2}} \dfrac{1}{u} - 7u^{-2} + 12u^{-3} - 6u^{-4} du = − ( ln ⁡ ( u ) + 7 u − 6 u 2 + 2 u 3 ) ∣ 1 9 − 33 2 = -(\ln(u) + \dfrac{7}{u} - \dfrac{6}{u^2} + \dfrac{2}{u^3})|_{1}^{\frac{9 - \sqrt{33}}{2}} = − ( ln ⁡ ( 33 − 9 2 ) + 83 33 − 765 576 ) = -(\ln(\dfrac{\sqrt{33} - 9}{2}) + \dfrac{83\sqrt{33} - 765}{576}) = 765 − 83 33 576 − ln ⁡ ( 9 − 33 2 ) ≈ 0.01317003 \dfrac{765 - 83\sqrt{33}}{576} - \ln(\dfrac{9 - \sqrt{33}}{2}) \approx \boxed{0.01317003}

and,

∫ − 7 + 33 2 0 f ( x ) d x = ∫ 1 9 − 33 2 u 4 − 15 u 3 + 50 u 2 − 60 u + 24 u 5 d u = ∫ 1 9 − 33 2 1 u − 15 u − 2 + 50 u − 3 − 60 u − 4 + 24 u − 5 d u = \int_{\frac{-7 + \sqrt{33}}{2}}^{0} f(x) dx = \int_{1}^{\frac{9 - \sqrt{33}}{2}} \dfrac{u^4 - 15u^3 + 50u^2 - 60u + 24}{u^5} du = \int_{1}^{\frac{9 - \sqrt{33}}{2}} \dfrac{1}{u} - 15u^{-2} + 50u^{-3} - 60u^{-4} + 24u^{-5} du = ( ln ⁡ ( u ) + 15 u − 25 u 2 + 20 u 3 − 6 u 4 ) ∣ 1 9 − 33 2 = (\ln(u) + \dfrac{15}{u} - \dfrac{25}{u^2} + \dfrac{20}{u^3} - \dfrac{6}{u^4})|_{1}^{\frac{9 - \sqrt{33}}{2}} = ln ⁡ ( 9 − 33 2 ) + 6087776403456 + 396474974208 33 2348273369088 − 4 \ln(\dfrac{9 - \sqrt{33}}{2}) + \dfrac{6087776403456 + 396474974208\sqrt{33}}{2348273369088} - 4 ≈ 0.04952097 \approx \boxed{0.04952097}

⟹ ∫ − 7 + 33 2 0 f ( x ) − g ( x ) d x = 0.03635094 \implies \int_{\frac{-7 + \sqrt{33}}{2}}^{0} f(x) - g(x) dx = \boxed{0.03635094}

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Your problems tend to be extremely computational, sir. Perhaps making the computation less heavy would make the problems more approachable (I like the idea behind your problems but once I see the computaiton, i get discouraged) Thanks.

D S - 3 years ago

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Thank you. Your right, at times I get carried away. In the future I'll attempt to make my problems less computational.

Rocco Dalto - 3 years ago

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