Limit $f(x)$ Evaluation
The function is defined as \( f(x) = \lim_{n \to \infty} \frac{1}{n^3} \sum_{k=1}^n \left[ \frac{k^2}{3^x} \right] \). We use the inequality \( y - 1 < [y] \le y \).
Applying this, we get bounds for the term inside the limit:
$ \frac{1}{n^3} \sum_{k=1}^n \left( \frac{k^2}{3^x} - 1 \right) < \frac{1}{n^3} \sum_{k=1}^n \left[ \frac{k^2}{3^x} \right] \le \frac{1}{n^3} \sum_{k=1}^n \frac{k^2}{3^x} $We evaluate the limits of the bounds as \( n \to \infty \). Using the formula \( \sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6} \), we find the limit:
$ \lim_{n \to \infty} \frac{1}{n^3} \sum_{k=1}^n \frac{k^2}{3^x} = \frac{1}{3^x} \lim_{n \to \infty} \frac{n(n+1)(2n+1)}{6n^3} $ $ = \frac{1}{3^x} \cdot \frac{1}{6} \lim_{n \to \infty} \frac{n^3(1+1/n)(2+1/n)}{n^3} = \frac{1}{3^x} \cdot \frac{1}{6} \cdot 2 = \frac{1}{3 \cdot 3^x} = \frac{1}{3^{x+1}} $The limit of the lower bound is also \( \frac{1}{3^{x+1}} \). By the Squeeze Theorem, \( f(x) = \frac{1}{3^{x+1}} \).
Summation $\sum f(j)$ Calculation
We need to compute the sum $S = \sum_{j=1}^\infty f(j)$:
$ S = \sum_{j=1}^\infty \frac{1}{3^{j+1}} $This is an infinite geometric series with the first term \( a = \frac{1}{3^{1+1}} = \frac{1}{9} \) and the common ratio \( r = \frac{1}{3} \).
The sum of an infinite geometric series is given by \( S = \frac{a}{1-r} \) (where \( |r| < 1 \)).
$ S = \frac{1/9}{1 - 1/3} = \frac{1/9}{2/3} = \frac{1}{9} \times \frac{3}{2} = \frac{1}{6} $Sum $12 \sum f(j)$
Finally, we calculate the required value:
$ 12 \times S = 12 \times \frac{1}{6} = 2 $The result is 2.
Similar Questions
- If $\int (\sin x)^{\frac{-11}{2}} (\cos x)^{\frac{-5}{2}} dx = - \frac{p_1}{q_1} (\cot x)^{\frac{9}{2}} - \frac{p_2}{q_2} (\cot x)^{\frac{5}{2}} - \frac{p_3}{q_3} (\cot x)^{\frac{1}{2}} + \frac{p_4}{q_4} (\cot x)^{\frac{-3}{2}} + C$, where $p_i$ and $q_i$ are positive integers with $\gcd(p_i, q_i) = 1$ for $i = 1, 2, 3, 4$ and C is the constant of integration, then $\frac{15 p_1 p_2 p_3 p_4}{q_1 q_2 q_3 q_4}$ is equal to _________
- Let a differentiable function $f$ satisfy the equation $\int_{0}^{36} f\left(\frac{tx}{36}\right) dt = 4\alpha f(x)$. If $y = f(x)$ is a standard parabola passing through the points (2, 1) and (– 4, $\beta$), then $\beta^\alpha$ is equal to ______.
- Let $f(\alpha)$ denote the area of the region in the first quadrant bounded by $x = 0, x = 1, y^2 = x$ and $y = |\alpha x - 5| - |1 - \alpha x| + \alpha x^2$. Then $(f(0) + f(1))$ is equal to
Let $f(x) = \int \frac{7x^{10} + 9x^8}{(1 + x^2 + 2x^9)^2} \,dx$, $x > 0$, $\lim_{x \rightarrow 0} f(x) = 0$ and $f(1) = \frac{1}{4}$.
If $A = \begin{bmatrix} 0 & 0 & 1 \\ \frac{1}{4} & f'(1) & 1 \\ \alpha^2 & 4 & 1 \end{bmatrix}$ and $B = \text{adj}(\text{adj } A)$ be such that $|B| = 81$, then $\alpha^2$ is equal to
Let $f$ be a differentiable function satisfying $f(x) = 1 - 2x + \int_{0}^{x} e^{(x - t)} f(t) dt$, $x \in \mathbf{R}$ and let $g(x) = \int_{0}^{x} (f(t) + 2)^{15} (t - 4)^{6} (t + 12)^{17} dt$, $x \in \mathbf{R}$. If $\text{p}$ and $\text{q}$ are respectively the points of local minima and local maxima of $g$, then the value of $|\text{p} + \text{q}|$ is equal to _________.
- Let $f : \mathbf{R} \rightarrow \mathbf{R}$ be a function such that $f(x) + 3f\left(\frac{\pi}{2} - x\right) = \sin x$, $x \in \mathbf{R}$. Let the maximum value of $f$ on $\mathbf{R}$ be $\alpha$. If the area of the region bounded by the curves $g(x) = x^{2}$ and $h(x) = \beta x^{3}, \beta > 0$, is $\alpha^{2}$, then $30\beta^{3}$ is equal to ___________.
Let $f$ be a twice differentiable function such that $f(x) = \int_{0}^{x} \tan(t - x)dt - \int_{0}^{x} f(t)\tan t dt$, $x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.
Then $f''\left(\frac{\pi}{6}\right) + 12f'\left(-\frac{\pi}{6}\right) + f\left(\frac{\pi}{6}\right)$ is equal to __________- Let $y = y(x)$ be the solution of the differential equation $\left(x^2 - x\sqrt{x^2 - 1}\right)dy + \left(y\left(x - \sqrt{x^2 - 1}\right) - x\right)dx = 0, x \geq 1$. If $y(1) = 1$, then the greatest integer less than $y\left(\sqrt{5}\right)$ is _________.
If $\int \frac{2x+5}{\sqrt{7-6x-x^2}} dx$ = $A\sqrt{7-6x-x^2} + Bsin^{-1} \left( \frac{x+3}{4} \right) + C$
(Where C is a constant of integration), then the ordered pair (A,B) is equal to :-
- If $I_1 = \int_0^1 e^{-x} cos^2x dx$, $I_2 = \int_0^1 e^{-x^2} cos^2x dx$ and $I_3 = \int_0^1 e^{-x^2} dx$; then :
Important Questions from Integral Calculus
- If $\int (\sin x)^{\frac{-11}{2}} (\cos x)^{\frac{-5}{2}} dx = - \frac{p_1}{q_1} (\cot x)^{\frac{9}{2}} - \frac{p_2}{q_2} (\cot x)^{\frac{5}{2}} - \frac{p_3}{q_3} (\cot x)^{\frac{1}{2}} + \frac{p_4}{q_4} (\cot x)^{\frac{-3}{2}} + C$, where $p_i$ and $q_i$ are positive integers with $\gcd(p_i, q_i) = 1$ for $i = 1, 2, 3, 4$ and C is the constant of integration, then $\frac{15 p_1 p_2 p_3 p_4}{q_1 q_2 q_3 q_4}$ is equal to _________
- Let a differentiable function $f$ satisfy the equation $\int_{0}^{36} f\left(\frac{tx}{36}\right) dt = 4\alpha f(x)$. If $y = f(x)$ is a standard parabola passing through the points (2, 1) and (– 4, $\beta$), then $\beta^\alpha$ is equal to ______.
- Let $f(\alpha)$ denote the area of the region in the first quadrant bounded by $x = 0, x = 1, y^2 = x$ and $y = |\alpha x - 5| - |1 - \alpha x| + \alpha x^2$. Then $(f(0) + f(1))$ is equal to
Let $f(x) = \int \frac{7x^{10} + 9x^8}{(1 + x^2 + 2x^9)^2} \,dx$, $x > 0$, $\lim_{x \rightarrow 0} f(x) = 0$ and $f(1) = \frac{1}{4}$.
If $A = \begin{bmatrix} 0 & 0 & 1 \\ \frac{1}{4} & f'(1) & 1 \\ \alpha^2 & 4 & 1 \end{bmatrix}$ and $B = \text{adj}(\text{adj } A)$ be such that $|B| = 81$, then $\alpha^2$ is equal to
Let $f$ be a differentiable function satisfying $f(x) = 1 - 2x + \int_{0}^{x} e^{(x - t)} f(t) dt$, $x \in \mathbf{R}$ and let $g(x) = \int_{0}^{x} (f(t) + 2)^{15} (t - 4)^{6} (t + 12)^{17} dt$, $x \in \mathbf{R}$. If $\text{p}$ and $\text{q}$ are respectively the points of local minima and local maxima of $g$, then the value of $|\text{p} + \text{q}|$ is equal to _________.
