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harina [27]
3 years ago
11

10 POINTS!!! FULL ANSWER IN STEP BY STEP FORMAT!!

Mathematics
1 answer:
In-s [12.5K]3 years ago
7 0
Remark
This question likely should be done before the other one. What you are trying to do is give C a value. So you need to remember that C is always part of an indefinite integral.

y = \int (cos(x) + sin(x) ) \, dx = \int{cos(x) \,dx + \int sin(x) \,dx
y = sin(x)  -  cos(x) + C 
y(π) = sin(π) - cos(π) + C = 0
y(π) = 0  -(-1) + C = 0
y(π) = 1 + C = 0
C = - 1

y = sin(x) - cos(x) - 1  <<<<< Answer

Problem Two
Remember that y =  \int\ { \frac{1}{x} } \, dx = ln(|x|) + C
y( - e^3 ) = ln(|x|) + C = 0
y(-e^3) = ln(|-e^3|) + C = 0
y(-e^3) = 3 + C  = 0
3 + C = 0
C = - 3

y = ln(|x|) - 3  <<<< Answer
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