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# MTH301 Calculus 2, GDB # 5, Opening Date August 12, 2015, Closing Date August 18, 2015

Topic : Line Integral

Question:

Grading criteria will be as follows:

1. If substitution done correctly then you will get  25% marks
2. In addition if limits correctly taken then you will get 50% marks
3. If integration is done correctly then you will get 75% marks
4. If after simplification answer is correct then you will get  100% marks

Note: It is mandatory to mention all necessary calculation steps to get full marks.

• Write the mathematical equations/symbols in Math Type software and then copy at GDB. For the procedure watch this video based on the problems and solution of previous GDB  Problem and Solutions Video.
• Last date of this GDB is August 17, 2015. There will not any relaxation after due date. You have ample days to complete this task. Submit your answers at the earliest to avoid losing your marks.

Important Note: A new functionality (Preview) has been added in GDB to verify that your post is correctly displaying. So before posting your comments you should first preview the post and if the equations are correctly displaying then you can post the comment. Remember once you post the comments on GDB then you cannot edit or repost your comment.

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### Replies to This Discussion

best of luck all students

Koi to kuch batao, last date ka w8 q kr rahy ho sb. Discussion start kro guyz.

here is the first part of gdb ...plzzzzzzz check and tel me that is right or not???

$Solution:$$= \int\limits_c {({x^2}y\;dx + y\;dy)..............(i)}$$Let$$P = {x^2}y\;dx$$Q = y\;dy$$Putting\;the\;value\;of\;y\;in\;P\;and\;Q$$P = {x^2}({x^2})\;dx = {x^4}\;dx$$Q = {x^2}\;dy$$y = {x^2}$$\frac{{dy}}{{dx}} = \frac{d}{{dx}}({x^2})$$dy = 2x\;dx$$Putting\;the\;value\;in\;equation\;(i)$$= \int\limits_c {{x^4}dx + {x^2}dy}$$= \int\limits_c {{x^4}dx + {x^2}(2x)dx}$$= \int\limits_c {{x^4}dx + 2{x^3}dx}$
koi repaly hi kar do

$\int {{x^2}y{\text{ }}dx} + y{\text{ }}dy...........(a)$$let$$P = {x^2}y{\text{ }}dx$$Q = y{\text{ }}dy$${\text{Now Putting the value of y in p and q }}$$P = {x^2}({x^2}) = {x^4}$$and$$Q = {x^2}dy$$Now,$$y = {x^2}$$\frac{{dy}}{{dx}} = 2x$$dy = 2x{\text{ }}dx$${\text{putting values in equation (a)}}{\text{.}}$$= \int {{x^4}{\text{ }}dx} + 2{x^3}{\text{ }}dx$${\text{limit should be applied that is (}}0{\text{,2)}}{\text{.}}$$= {}_0^2\int {({x^4}{\text{ }}dx} + 2{x^3}{\text{ }}dx)$${\text{integrate w}}{\text{.r}}{\text{.t x}}{\text{.}}$$= \left| {\frac{{{x^5}}}{5}} \right|_0^2 + 2\left| {\frac{{{x^4}}}{4}} \right|_0^2$$= \frac{{32}}{5} + \frac{{16}}{2}$$= \frac{{64 + 80}}{{10}}$$= \frac{{144}}{{10}}$$= \frac{{72}}{5}{\text{ Answer}}{\text{.}}$

Plz tell me apna ye solution kese post kia ha..??ma kr raha hn to error dy raha ha

yahooo

thanks for replaying

anee jo ap na jo lms ma post kya hai wo copy kr k idhr post kr dangi .. coz me apni post kr rha hoo tu error da rha .. or ap email krdan... Bc140201235

pleas any one tell me this the write solution

here is the full and right solution of gdb .............enjoy remember me in payers

$\begin{gathered} \hfill \\ gdb\,\,\,\,Solution: \hfill \\ = \int\limits_c {({x^2}y\;dx + y\;dy)..............(i} ) \hfill \\ Let \hfill \\ P = {x^2}y\;dx \hfill \\ Q = y\;dy \hfill \\ give\;the\;value\;of\;y\;to\;P\;and\;Q \hfill \\ P = {x^2}({x^2})\;dx = {x^4}\;dx \hfill \\ Q = {x^2}\;dy \hfill \\ y = {x^2} \hfill \\ \frac{{dy}}{{dx}} = \frac{d}{{dx}}({x^2}) \hfill \\ dy = 2x\;dx \hfill \\ give\;the\;value\;to\;equation\;.....(i) \hfill \\ = \int\limits_c {{x^4}dx + {x^2}dy} \hfill \\ = \int\limits_c {{x^4}dx + {x^2}(2x)dx} \hfill \\ = \int\limits_c {{x^4}dx + 2{x^3}dx} \hfill \\ Limit\;that\;applied\,is\;(0,2) \hfill \\ = \int\limits_0^2 {({x^4}dx + 2{x^3}dx)} \hfill \\ Integrate\;w.r.t\;x \hfill \\ = \left| {\left. {\frac{{{x^5}}}{5}} \right|} \right._0^2 + 2\left| {\frac{{{x^4}}}{4}} \right|_0^2 \hfill \\ = \left[ {\frac{{{{(2)}^5}}}{5} - \frac{{{{(0)}^5}}}{5}} \right] + 2\left[ {\frac{{{{(2)}^4}}}{4} - \frac{{{{(0)}^4}}}{4}} \right] \hfill \\ = \frac{{32}}{5} + \frac{{16}}{2} \hfill \\ = \frac{{32}}{5} + \frac{{16}}{2} \hfill \\ = \frac{{64 + 80}}{{10}} \hfill \\ = \frac{{144}}{{10}} \hfill \\ = \frac{{72}}{5} \hfill \\ = 14.4 \hfill \\ Answer \hfill \\ \end{gathered}$$\begin{gathered} \hfill \\ gdb\,\,\,\,Solution: \hfill \\ = \int\limits_c {({x^2}y\;dx + y\;dy)..............(i} ) \hfill \\ Let \hfill \\ P = {x^2}y\;dx \hfill \\ Q = y\;dy \hfill \\ give\;the\;value\;of\;y\;to\;P\;and\;Q \hfill \\ P = {x^2}({x^2})\;dx = {x^4}\;dx \hfill \\ Q = {x^2}\;dy \hfill \\ y = {x^2} \hfill \\ \frac{{dy}}{{dx}} = \frac{d}{{dx}}({x^2}) \hfill \\ dy = 2x\;dx \hfill \\ give\;the\;value\;to\;equation\;.....(i) \hfill \\ = \int\limits_c {{x^4}dx + {x^2}dy} \hfill \\ = \int\limits_c {{x^4}dx + {x^2}(2x)dx} \hfill \\ = \int\limits_c {{x^4}dx + 2{x^3}dx} \hfill \\ Limit\;that\;applied\,is\;(0,2) \hfill \\ = \int\limits_0^2 {({x^4}dx + 2{x^3}dx)} \hfill \\ Integrate\;w.r.t\;x \hfill \\ = \left| {\left. {\frac{{{x^5}}}{5}} \right|} \right._0^2 + 2\left| {\frac{{{x^4}}}{4}} \right|_0^2 \hfill \\ = \left[ {\frac{{{{(2)}^5}}}{5} - \frac{{{{(0)}^5}}}{5}} \right] + 2\left[ {\frac{{{{(2)}^4}}}{4} - \frac{{{{(0)}^4}}}{4}} \right] \hfill \\ = \frac{{32}}{5} + \frac{{16}}{2} \hfill \\ = \frac{{32}}{5} + \frac{{16}}{2} \hfill \\ = \frac{{64 + 80}}{{10}} \hfill \\ = \frac{{144}}{{10}} \hfill \\ = \frac{{72}}{5} \hfill \\ = 14.4 \hfill \\ Answer \hfill \\ \end{gathered}$

thnks alot

Mani BSc apke waja Bhot bata masla ban gya Line Brack nhe dale apne pleas agar ata nhe tho copy past kar ka logo ke prysani ke waja na bano

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