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MTH301 - Assignment No' 3 - Due Date: 27th July 2017

Regarding Assignment # 3 Dated: Jul 21, 17
Dear Students,
Assignment No.3 has been uploaded on VULMS. Its due date is 27th July 2017. Solve your Assignment in ms word and properly upload it through VULMS with in the due time. No assignment will be accepted after the due date.

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guys rat ko 9 bje tk confrm ho ga kon sa solution thik h sb us time ka wait kr lo phir sb send kr len ge

 please somebody share correct solution.

      kam az kam office walon pr bhe raham kia karo Zaaleemoo

MTH301 Assignment#03 Solution

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Assigmnent#3 Soluion File of MTH301

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attached complete solution..

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Where the solution file

QNo.1 Solution:
Evaluate ∫ [(0,0) to (4,3)] (3x²e^y) dx + (x³ e^y) dy where the given line integral is independent of the path. 

—————————————————————————————— 

Since line integral is independent of path, we can choose the simplest path possible: a straight line from (0,0) to (4,3) 
x = 4t ----> dx = 4 dt 
y = 3t ----> dy = 3 dt 
t = 0 to 1 

∫ [(0,0) to (4,3)] (3x²e^y) dx + (x³ e^y) dy 
= ∫ [0 to 1] (3(4t)²e^(3t)) (4 dt) + ((4t)³ e^(3t)) (3 dt) 
= ∫ [0 to 1] 192 (t² e^(3t) + t³ e^(3t)) dt 
= 64e³ 
≈ 1285.5 

You can use integration by parts to find result above. 
I'll leave that to you. 

—————————————————————————————— 

However, we can find a path that will make calculating integral simpler: 
Path from (0,0) to (0,3) to (4,3) 

(0,0) to (0,3) 
x = 0 ----> dx = 0 dt 
y = 3t ---> dy = 3 dt 

∫ [(0,0) to (0,3)] (3x²e^y) dx + (x³ e^y) dy 
= ∫ [0 to 1] (0) (0 dt) + (0) 3 dt 
= 0 

(0,3) to (4,3) 
x = 4t ----> dx = 4 dt 
y = 3 -----> dy = 0 dt 

∫ [(0,3) to (4,3)] (3x²e^y) dx + (x³ e^y) dy 
= ∫ [0 to 1] (3x(4t)²e^3) (4 dt) + ((4t)³ e^3) (0 dt) 
= ∫ [0 to 1] (192e³) t² dt 
= 64e³ t³ | [0 to 1] 
= 64e³ 

Add results from both paths: 

∫ [(0,3) to (4,3)] (3x²e^y) dx + (x³ e^y) dy 
= 0 + 64e³ 
= 64e³

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