Arithmetic-Geometric Series App 2.

Calculus Level 4

Let ∣ x ∣ < 1 |x| < 1 and f ( x ) = ∑ n = 0 ∞ ( a + n d ) x n f(x) = \sum_{n = 0}^{\infty} (a + nd)x^n and g ( x ) = ∑ n = 0 ∞ ( a ∗ + n 2 d ∗ ) x n g(x) = \sum_{n = 0}^{\infty} (a^{*} + n^2 d^{*})x^n and f ( 1 2 ) = g ( 1 2 ) = 2 f(\dfrac{1}{2}) = g(\dfrac{1}{2}) = 2 and f ( − 1 2 ) = g ( − 1 2 ) = 2 9 f(-\dfrac{1}{2}) = g(-\dfrac{1}{2}) = \dfrac{2}{9} .

The volume V V formed when the region bounded by f ( x ) f(x) and g ( x ) g(x) on [ − 1 2 , 1 2 ] [-\dfrac{1}{2},\dfrac{1}{2}] is revolved about the x x axis can be expressed as V = α β π V = \dfrac{\alpha}{\beta}\pi , where α \alpha and β \beta are coprime positive integers.

Find α + β \alpha + \beta .

Refer To Previous Problem


The answer is 23813.

Rocco Dalto by Rocco Dalto , 67, USA

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1 solution

Rocco Dalto
Aug 3, 2018

Let ∣ x ∣ < 1 |x| < 1 .

f ( x ) = ∑ n = 0 ∞ ( a + n d ) x n = a ∑ n = 0 ∞ x n + d ∑ n = 1 ∞ n x n = f(x) = \sum_{n = 0}^{\infty} (a + nd)x^n = a\sum_{n = 0}^{\infty} x^n + d\sum_{n = 1}^{\infty} nx^n = a 1 − x + d ∑ j = 1 ∞ ∑ n = j ∞ x n = \dfrac{a}{1 - x} + d\sum_{j = 1}^{\infty}\sum_{n = j}^{\infty} x^n = a 1 − x + d 1 − x ∑ j = 1 ∞ x j = a 1 − x + d 1 − x ( x 1 − x ) = a 1 − x + d x ( 1 − x ) 2 \dfrac{a}{1 - x} + \dfrac{d}{1 - x}\sum_{j = 1}^{\infty} x^{j} = \dfrac{a}{1 - x} + \dfrac{d}{1 - x}(\dfrac{x}{1 - x}) = \boxed{\dfrac{a}{1 - x} + \dfrac{dx}{(1 - x)^2}}

and,

g ( x ) = ∑ n = 0 ∞ ( a ∗ + n 2 d ∗ ) x n = a ∗ ∑ n = 0 ∞ x n + d ∗ ∑ n = 1 ∞ n 2 x n = a ∗ 1 − x + d ∗ ∑ j = 1 ∞ ∑ n = j ∞ n x n = g(x) =\sum_{n = 0}^{\infty} (a^{*} + n^2 d^{*})x^n = a^{*}\sum_{n = 0}^{\infty} x^n + d^{*}\sum_{n = 1}^{\infty} n^2 x^n = \dfrac{a^{*}}{1 - x} + d^{*}\sum_{j = 1}^{\infty}\sum_{n = j}^{\infty} n x^n = ( a ∗ 1 − x + d ∗ ∑ j = 1 ∞ ( j x j + ( j + 1 ) x j + 1 + ( j + 2 ) x j + 2 + . . . ) = (\dfrac{a^{*}}{1 - x} + d^{*}\sum_{j = 1}^{\infty} (j x^{j} + (j + 1)x^{j + 1} + (j + 2)x^{j + 2} + ... ) = a ∗ 1 − x + d ∗ ( 1 1 − x ∑ j = 1 ∞ j x j + x ( 1 − x ) 2 ∑ j = 1 ∞ x j ) = \dfrac{a^{*}}{1 - x} + d^{*}(\dfrac{1}{1 - x}\sum_{j = 1}^{\infty} j x^{j} + \dfrac{x}{(1 - x)^2}\sum_{j = 1}^{\infty} x^{j}) = a ∗ 1 − x + d ∗ ( ( 1 1 − x ) ( x ( 1 − x ) 2 ) + ( x ( 1 − x ) 2 ) ( x 1 − x ) ) = a ∗ 1 − x + d ∗ ( x 2 + x ( 1 − x ) 3 ) \dfrac{a^{*}}{1 - x} + d^{*}((\dfrac{1}{1 - x})(\dfrac{x}{(1 - x)^2}) + (\dfrac{x}{(1 - x)^2})(\dfrac{x}{1 - x})) = \boxed{\dfrac{a^{*}}{1 - x} + d^{*}(\dfrac{x^2 + x}{(1 - x)^3})}

f ( 1 2 ) = 2 f(\dfrac{1}{2}) = 2 and f ( − 1 2 ) = 2 9 ⟹ f(-\dfrac{1}{2}) = \dfrac{2}{9} \implies

3 a − d = 1 3a - d = 1

a + d = 1 a + d = 1

⟹ a = 1 2 = d ⟹ f ( x ) = 1 2 ( 1 ( 1 − x ) 2 ) ⟹ ∫ − 1 2 1 2 ( f ( x ) ) 2 d x = 1 12 ( 1 ( 1 − x ) 3 ) ∣ − 1 2 1 2 = 52 81 \implies a = \dfrac{1}{2} = d \implies f(x) = \dfrac{1}{2}(\dfrac{1}{(1 - x)^2}) \implies \displaystyle\int_{-\frac{1}{2}}^{\frac{1}{2}} (f(x))^2 dx = \dfrac{1}{12}(\dfrac{1}{(1 - x)^3})|_{-\frac{1}{2}}^{\frac{1}{2}} = \boxed{\dfrac{52}{81}} .

g ( 1 2 ) = 2 g(\dfrac{1}{2}) = 2 and g ( − 1 2 ) = 2 9 ⟹ g(-\dfrac{1}{2}) = \dfrac{2}{9} \implies

9 a ∗ − d ∗ = 3 9a^{*} - d^{*} = 3

a ∗ + 3 d ∗ = 1 a^{*} + 3d^{*} = 1

⟹ a ∗ = 5 14 \implies a^{*} = \dfrac{5}{14} and d ∗ = 3 14 d^{*} = \dfrac{3}{14} ⟹ g ( x ) = 1 14 ( 5 1 − x + 3 ( x 2 + x ) ( 1 − x ) 3 ) \implies g(x) = \dfrac{1}{14}(\dfrac{5}{1 - x} + \dfrac{3(x^2 + x)}{(1 - x)^3})

∫ − 1 2 1 2 ( g ( x ) ) 2 d x = 1 1 4 2 ∫ − 1 2 1 2 25 ( 1 − x ) 2 + 30 x ( x + 1 ) ( 1 − x ) 4 + 9 x 2 ( x + 1 ) 2 ( 1 − x ) 6 d x \displaystyle\int_{-\frac{1}{2}}^{\frac{1}{2}} (g(x))^2 dx = \dfrac{1}{14^2}\displaystyle\int_{-\frac{1}{2}}^{\frac{1}{2}} \dfrac{25}{(1 - x)^2} + \dfrac{30x(x + 1)}{(1 - x)^4} + \dfrac{9x^2(x + 1)^2}{(1 - x)^6} dx

Let I 1 ( x ) = ∫ x ( x + 1 ) ( 1 − x ) 4 d x I_{1}(x) = \displaystyle\int \dfrac{x(x + 1)}{(1 - x)^4} dx

Letting u = 1 − x ⟹ d x − − d u ⟹ I 1 ( x ) = − ∫ u − 2 − 3 u − 3 + 2 u − 4 d u = 1 1 − x − 3 2 ( 1 − x ) 2 + 2 3 ( 1 − x ) 3 u = 1 - x \implies dx - -du \implies I_{1}(x) = -\displaystyle\int u^{-2} - 3u^{-3} + 2u^{-4} du = \dfrac{1}{1 - x} - \dfrac{3}{2(1 - x)^2} + \dfrac{2}{3(1 - x)^3}

⟹ I 1 ( x ) ∣ − 1 2 1 2 = 92 81 \implies I_{1}(x)|_{-\frac{1}{2}}^{\frac{1}{2}} = \boxed{\dfrac{92}{81}}

Let I 2 ( x ) = ∫ x 2 ( x + 1 ) 2 ( 1 − x ) 6 d x I_{2}(x) = \displaystyle\int \dfrac{x^2(x + 1)^2}{(1 - x)^6} dx

Letting u = 1 − x ⟹ d x − − d u ⟹ I 2 = − ∫ u − 2 − 6 u − 3 + 13 u − 4 − 12 u − 5 + 4 u − 6 d u = u = 1 - x \implies dx - -du \implies I_{2} = -\displaystyle\int u^{-2} - 6u^{-3} + 13u^{-4} - 12u^{-5} + 4u^{-6} du = 1 1 − x − 3 ( 1 − x ) 2 + 13 3 ( 1 − x ) 3 − 3 ( 1 − x ) 4 + 4 5 ( 1 − x ) 5 \dfrac{1}{1 - x} - \dfrac{3}{(1 - x)^2} + \dfrac{13}{3(1 - x)^3} - \dfrac{3}{(1 - x)^4} + \dfrac{4}{5(1 - x)^5}

⟹ I 2 ( x ) ∣ − 1 2 1 2 = 2596 1215 \implies I_{2}(x)|_{-\frac{1}{2}}^{\frac{1}{2}} = \boxed{\dfrac{2596}{1215}}

and,

I 3 = ∫ − 1 2 1 2 ( 1 − x ) − 2 d x = 1 1 − x ∣ − 1 2 1 2 = 4 3 I_{3} = \displaystyle\int_{-\frac{1}{2}}^{\frac{1}{2}} (1 - x)^{-2} dx = \dfrac{1}{1 - x}|_{-\frac{1}{2}}^{\frac{1}{2}} = \boxed{\dfrac{4}{3}}

⟹ ∫ − 1 2 1 2 ( g ( x ) ) 2 = 1 1 4 2 ( 100 3 + 2760 81 + 23364 1215 ) = 2924 6615 \implies \displaystyle\int_{-\frac{1}{2}}^{\frac{1}{2}} (g(x))^2 = \dfrac{1}{14^2}(\dfrac{100}{3} + \dfrac{2760}{81} + \dfrac{23364}{1215}) = \boxed{\dfrac{2924}{6615}}

⟹ V = π ∫ − 1 2 1 2 ( f ( x ) ) 2 − ( g ( x ) ) 2 d x = π ( 52 81 − 2924 6615 ) = 3968 19845 π = α β π ⟹ α + β = 23813 \implies V = \pi\displaystyle\int_{-\frac{1}{2}}^{\frac{1}{2}} (f(x))^2 - (g(x))^2 dx = \pi(\dfrac{52}{81} - \dfrac{2924}{6615}) = \dfrac{3968}{19845}\pi = \dfrac{\alpha}{\beta}\pi \implies \alpha + \beta = \boxed{23813} .

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