Log Derivative Integral

Calculus Level 3

What is the sum of all the values of x x that satisfy log ⁡ x 2 { d d x ∫ x x + 1 1 2 ( t 2 − t ) d t } = x 2 − 8 x + 15 2 ? \displaystyle \log_{x^2} \left\{ \frac{d}{dx} \int_{x}^{x+1} \frac{1}{2}(t^2-t)\ dt \right\} = x^2-8x+\frac{15}{2} ?

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Mahindra Jain by Mahindra Jain

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1 solution

Tom Engelsman
Nov 7, 2020

If we calculate out the LHS first, we obtain:

∫ x x + 1 t 2 − t 2 d t = t 3 6 − t 2 4 ∣ x x + 1 = ( x + 1 ) 3 6 − ( x + 1 ) 2 4 − x 3 6 + x 2 4 \int_{x}^{x+1} \frac{t^2-t}{2} dt = \frac{t^3}{6} - \frac{t^2}{4}|_{x}^{x+1} = \frac{(x+1)^3}{6}-\frac{(x+1)^2}{4} - \frac{x^3}{6}+\frac{x^2}{4} ;

which differentiates & simplifies into ( x + 1 ) 2 2 − x + 1 2 − x 2 2 + x 2 = x ⇒ log ⁡ x 2 ( x ) = log ⁡ x 2 ( x 2 ) 1 / 2 = 1 2 . \frac{(x+1)^2}{2}-\frac{x+1}{2} - \frac{x^2}{2}+\frac{x}{2} = x \Rightarrow \log_{x^2}(x) = \log_{x^2} (x^2)^{1/2} = \frac{1}{2}.

We can now solve the quadratic equation 0 = x 2 − 8 x + 7 = ( x − 1 ) ( x − 7 ) ⇒ x = 1 , 7 0 = x^2 - 8x + 7 = (x-1)(x-7) \Rightarrow x = 1, 7 . Since the original logarithm on the LHS requires the base to be x 2 > 1 x^2 > 1 , the only permissible answer is x = 7 . \boxed{x=7}.

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