Rahul K.

Intern at VSSC,Thiruvananthapuram and Trainee at SAIL,Bokaro

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Basic Math

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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

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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

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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} \\

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