Asked by Kelly
Evaluate the definite integral:
sqrt(8-2x)
lower limit=-7
upper limit=0
I got -(1/3)(8-2x)^(3/2) and it was wrong. Please Help! Thanks in advance!
sqrt(8-2x)
lower limit=-7
upper limit=0
I got -(1/3)(8-2x)^(3/2) and it was wrong. Please Help! Thanks in advance!
Answers
Answered by
Jai
Actually your answer is correct if it's indefinite integral. :)
Now that you got the answer, we can now evaluate it at the given bounds:
-(1/3)(8-2x)^(3/2) , at x = -7 to 0
= -(1/3)(8-2(0))^(3/2) - [-(1/3)(8-2(-7))^(3/2)]
= -(1/3)(8)^(3/2) - [-(1/3)(22)^(3/2)]
= (22/3)*(22)^(1/2) - (16/3)*(2)^(1/2)
= 26.85
Hope this helps :3
Now that you got the answer, we can now evaluate it at the given bounds:
-(1/3)(8-2x)^(3/2) , at x = -7 to 0
= -(1/3)(8-2(0))^(3/2) - [-(1/3)(8-2(-7))^(3/2)]
= -(1/3)(8)^(3/2) - [-(1/3)(22)^(3/2)]
= (22/3)*(22)^(1/2) - (16/3)*(2)^(1/2)
= 26.85
Hope this helps :3
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