TutorMe homepage

SIGN IN

Start Free Trial

Rahul K.

Intern at VSSC,Thiruvananthapuram

Tutor Satisfaction Guarantee

Basic Math

TutorMe

Question:

\\ {Question :}\\ Find the value of following definite integration\\ $$\int_{0}^{1} (1-x^{2008})^\frac{1}{2007}-(1-x^{2007})^\frac{1}{2008} dx$$ \\

Rahul K.

Answer:

Let $(1-x^{2008})^\frac{1}{2007}$=$y$ \\ then, $x=(1-y^{2007})^\frac{1}{2008}$\\ Now,$$\int_{0}^{1} (1-x^{2008})^\frac{1}{2007}-(1-x^{2007})^\frac{1}{2008} dx$$\\ As sum inside integral can divided into two parts with change of variables of second part\\ implies $$\int_{0}^{1} (1-x^{2008})^\frac{1}{2007}dx-\int_{0}^{1}(1-y^{2007})^\frac{1}{2008} dy$$\\ Since both the expressions can be written in other variable ,after integration they will give same value so they cancel each other to give answer as zero. So,$$\int_{0}^{1} (1-x^{2008})^\frac{1}{2007}dx-\int_{0}^{1}(1-y^{2007})^\frac{1}{2008} dy=0$$\\ Hence,$$\int_{0}^{1} (1-x^{2008})^\frac{1}{2007}-(1-x^{2007})^\frac{1}{2008} dx=0$$

Electrical Engineering

TutorMe

Question:

\\ {Question: }\\ Find current in every branch of the circuit using Mesh current method for given circuit.\\ \begin{figure} \centering \includegraphics[scale=0.75]{qfig.png} \caption{Figure1} \end{figure} \\ ** How to insert picture of circuit here**

Rahul K.

Answer:

\\ {Solution: }\\ The first step in mesh current method is to identify loops within the circuit and the assign current direction in the loops. The choice of current is arbitrary, we have to take anticlockwise currents $I_{1}$ in loop 1 and $I_{2}$ in loop 2.\\ Next step is to label all voltage drop polarities across resistors according to assumed direction of current in the loop. The upstream end of the resistor is negative while downstream is positive. Then figure will become \\ \begin{figure} \centering \includegraphics[scale=0.75]{sfig1.png} \caption{Figure2} \end{figure} \\ Then,applying Kirchoff's voltage for loops , we get the eqautions for loops \\ Loop 1 \\ \begin{align} \begin{split} -28+2(I_{1}+I_{2})+4I_{1} &=0\\ \end{split} \end{align} \\After simplification, \begin{align} \begin{split} 6I_{1}+2I_{2}&=28 \\ \end{split} \end{align} \\ Loop 2 \begin{align} \begin{split} -2(I_{1}+I_{2})+7-I_{2} &=0\\ \end{split} \end{align} \\After simplification \begin{align} \begin{split} 2I_{1}+3I_{2}&=7 \\ \end{split} \end{align} \\ On simplifying for $I_{1}$ and $I_{1}$ ,we get\\ $I_{1}$=5A \\ $I_{2}$=-1A \\ Negative sign of $I_{2}$ shows that the assumed direction of current for loop 2 is opposite of what we have taken . So figure will be\\ \begin{figure} \centering \includegraphics[scale=0.75]{sfig2.png} \caption{Figure3} \end{figure} \\ Now after calculating current in each branch we will have \\ \begin{figure} \centering \includegraphics[scale=0.75]{sfig3.png} \caption{Figure4} \end{figure} \\ **How to insert picture of circuit here**

Physics

TutorMe

Question:

{Question: }\\ If a particle is moving and traveled half the distance with velocity $v_{0}$.The remaining part of distance was covered with velocity $v_{1}$ for halftime and velocity $v_{2}$ for other half time.Find the mean velocity of particle averaged over the whole time of the motion. \\

Rahul K.

Answer:

{Solution: }\\ Let $s$ be the total distance traveled by the particle and $t_{1}$ be the time to taken to cover the half distance.\\ After that, let $2t$ be the time to cover the rest half of distance such that time $t$ for velocity $v_{1}$ and time $t$ for velocity $v_{2}$.\\ Therefore, \begin{align} \begin{split} \frac{s}{2} &= v_{0}t_{1} \\ t_{1}&=\frac{s}{2v_{0}}\\ \end{split} \end{align} \\ and \begin{align} \begin{split} \frac{s}{2} &= (v_{1}+v_{2})t \\ t&=\frac{s}{2(v_{1}+v_{2})}\\ \end{split} \end{align} \\ For average velocity, \\ \begin{align} \begin{split} v_{avg} &=\frac{s}{t_{1}+2t}\\ &=\frac{s}{(s/2v_{0})+(s/(v_{1}+v_{2}))}\\ &=\frac{2v_{0}(v_{1}+v_{2})}{v_{1}+v_{2}+2v_{0}} \\ \end{split} \end{align} \\

Send a message explaining your

needs and Rahul will reply soon.

needs and Rahul will reply soon.

Contact Rahul

Ready now? Request a lesson.

Start Session

FAQs

What is a lesson?

A lesson is virtual lesson space on our platform where you and a tutor can communicate.
You'll have the option to communicate using video/audio as well as text chat.
You can also upload documents, edit papers in real time and use our cutting-edge virtual whiteboard.

How do I begin a lesson?

If the tutor is currently online, you can click the "Start Session" button above.
If they are offline, you can always send them a message to schedule a lesson.

Who are TutorMe tutors?

Many of our tutors are current college students or recent graduates of top-tier universities
like MIT, Harvard and USC.
TutorMe has thousands of top-quality tutors available to work with you.

Made in California

© 2019 TutorMe.com, Inc.