Question:
If \( F(x)=\int x(\log x)^{2} \, dx \) and \( F(e)=\frac{e^{2}}{4} \), then \( F(1) = \)
If \( F(x)=\int x(\log x)^{2} \, dx \) and \( F(e)=\frac{e^{2}}{4} \), then \( F(1) = \)
Show Hint
Since \( \log 1 = 0 \), substituting \( x = 1 \) into your terms causes all components containing log to vanish completely, leaving only the pure polynomial constant fraction behind.
Updated On: Jun 8, 2026
- \( 0 \)
- \( \frac{1}{4} \)
- \( \frac{1}{2} \)
- \( \frac{3\log(2)}{4} \)
Show Solution
Verified By Collegedunia
The Correct Option is B
Solution and Explanation
Concept:
We use the method of integration by parts (\( \int u \, dv = u v - \int v \, du \)) twice, selecting the logarithmic term as \( u \) according to the ILATE rule.
Step 1: First round of integration by parts.
Let \( u = (\log x)^2 \implies du = 2\log x \cdot \frac{1}{x} dx \). Let \( dv = x \, dx \implies v = \frac{x^2}{2} \). \[ F(x) = \frac{x^2}{2}(\log x)^2 - \int \frac{x^2}{2} \cdot \left(\frac{2\log x}{x}\right) dx = \frac{x^2}{2}(\log x)^2 - \int x\log x \, dx \]
Step 2: Second round of integration by parts.
For \( \int x\log x \, dx \), let \( u = \log x \implies du = \frac{1}{x} dx \), and \( dv = x \, dx \implies v = \frac{x^2}{2} \): \[ \int x\log x \, dx = \frac{x^2}{2}\log x - \int \frac{x^2}{2} \cdot \frac{1}{x} dx = \frac{x^2}{2}\log x - \frac{x^2}{4} \] Combining everything together: \[ F(x) = \frac{x^2}{2}(\log x)^2 - \frac{x^2}{2}\log x + \frac{x^2}{4} + C \]
Step 3: Using the boundary condition to find constant \( C \).
We are given \( F(e) = \frac{e^2}{4} \). Substitute \( x = e \): \[ \frac{e^2}{4} = \frac{e^2}{2}(1)^2 - \frac{e^2}{2}(1) + \frac{e^2}{4} + C \implies \frac{e^2}{4} = \frac{e^2}{4} + C \implies C = 0 \]
Step 4: Evaluating \( F(1) \).
Substitute \( x = 1 \) into the fully determined expression (noting that \( \log 1 = 0 \)): \[ F(1) = 0 - 0 + \frac{1^2}{4} = \frac{1}{4} \] This matches option (B) perfectly.
Step 1: First round of integration by parts.
Let \( u = (\log x)^2 \implies du = 2\log x \cdot \frac{1}{x} dx \). Let \( dv = x \, dx \implies v = \frac{x^2}{2} \). \[ F(x) = \frac{x^2}{2}(\log x)^2 - \int \frac{x^2}{2} \cdot \left(\frac{2\log x}{x}\right) dx = \frac{x^2}{2}(\log x)^2 - \int x\log x \, dx \]
Step 2: Second round of integration by parts.
For \( \int x\log x \, dx \), let \( u = \log x \implies du = \frac{1}{x} dx \), and \( dv = x \, dx \implies v = \frac{x^2}{2} \): \[ \int x\log x \, dx = \frac{x^2}{2}\log x - \int \frac{x^2}{2} \cdot \frac{1}{x} dx = \frac{x^2}{2}\log x - \frac{x^2}{4} \] Combining everything together: \[ F(x) = \frac{x^2}{2}(\log x)^2 - \frac{x^2}{2}\log x + \frac{x^2}{4} + C \]
Step 3: Using the boundary condition to find constant \( C \).
We are given \( F(e) = \frac{e^2}{4} \). Substitute \( x = e \): \[ \frac{e^2}{4} = \frac{e^2}{2}(1)^2 - \frac{e^2}{2}(1) + \frac{e^2}{4} + C \implies \frac{e^2}{4} = \frac{e^2}{4} + C \implies C = 0 \]
Step 4: Evaluating \( F(1) \).
Substitute \( x = 1 \) into the fully determined expression (noting that \( \log 1 = 0 \)): \[ F(1) = 0 - 0 + \frac{1^2}{4} = \frac{1}{4} \] This matches option (B) perfectly.
Was this answer helpful?
0
0
Top AP EAPCET Mathematics Questions
- $D = \left\{x\in\mathbb{R}: f\left(x\right) =\sqrt{\frac{x - \left|x\right|}{x - \left[x\right]}} \text{is defined} \right\}$ and $C$ be the range of the real function $g(x) = \frac{2x}{4 + x^{2}}$. Then $D \cap C$
- $f\left(x\right) = \frac{x}{e^{x}-1} + \frac{x}{2} + 2 \cos^{3} \frac{x}{2} $ on $R-\left\{0\right\} $ is
- Consider the following lists.
List-I List-II (A) $f(x) = \frac{|x+2|}{x+2} , x \ne -2 $ (I) $[\frac{1}{3} , 1 ]$ (B) $(x)=|[x]|,x \in [R$ (II) Z (C) $h(x) = |x - [x]| , x \in [R$ (III) W (D) $f(x) = \frac{1}{2 - \sin 3x} , x \in [R$ (IV) [0, 1) (V) { -1, 1} - If $[x]$ is the greatest integer less than or equal to $x$ and $|x|$ is the modulus of $x$. then the system of three equations $2x + 3 | y | + 5[z] = 0, x + |y| - 2[z] = 4, x + |y| + |z| = 1$ has
- Investigate the values of $\lambda$ and $\mu$ for the system $x+2 y+3 z=6, x+3 y+5 z=9$, $2 x+5 y+\lambda z=\mu$ and match the values in List - I with the items in List - II.
List I List II (A) $\lambda=8, \mu \neq 15$ 1. Infinitely many solutions (B) $\lambda \neq 8, \mu \in R$ 2. No solution (C) $\lambda=8, \mu=15$ 3. Unique solution
View More Questions
Top AP EAPCET Integration Questions
- If \[ a^2 x^4 + b^2 y^4 = c^6, \] then the maximum value of \( xy \) is:
- Evaluate the integral \[ \int \frac{\sin^6 x}{\cos^8 x} \, dx. \]
- Evaluate the integral \[ \int \frac{x^5}{x^2 + 1} dx. \]
- Evaluate the integral \[ \int \sum_{r=0}^{\infty} \frac{x^r 3^r}{2r} dx. \]
- Evaluate the integral \[ \int \frac{x^4 + 1}{x^6 + 1} dx. \]
View More Questions
Top AP EAPCET Questions
- A body is projected from the ground at an angle of $\tan^{-1} \left( \frac{8}{7}\right)$ with the horizontal. The ratio of the maximum height attained by it to its range is
- The four quantum numbers of the valence electron of potassium are :
- A lens forms real and virtual images of an object, when the object is at $u_{1}$ and $u_{2}$ distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (take, the magnification of the real image as $m$ )
- Two spheres $P$ and $Q$, each of mass $200\, g$ are attached to a string of length one metre as shown in the figure. The string and the spheres are then whirled in a horizontal circle about $O$ at a constant angular speed. The ratio of the tension in the string between $P$ and $Q$ to that of between $P$ and $O$ is $(P$ is at mid-point of the line joining $O$ and $Q$ )
- The volume of $0.02\, M$ acidified permanganate solution required for complete reaction of $60\, mL$ of $0.01\, MI^{-}$ ion solution to form $I_2$ in $mL$ is
View More Questions