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Definite integral help again

Okay, I posted these two problems the other day and I thought I understood based on the help given, but I still don't really, so if anyone can give me a step by step on either one of them, or both then I'll be forever grateful!

1)

2)



Thank you thank you thank you if you can help.
(edited 10 years ago)
Reply 1
Original post by angelooo
Okay, I posted these two problems the other day and I thought I understood based on the help given, but I still don't really, so if anyone can give me a step by step on either one of them, or both then I'll be forever grateful!

1) 01cosπx2dx\displaystyle \int_{0}^{1}cos\frac{\pi x}{2}dx

2) 03(x+1)1/2\displaystyle \int_{0}^{3}({x+1})^{1/2}

Thank you thank you thank you if you can help.


ftfy
Reply 2
Original post by TenOfThem
ftfy

Thanks, haha, I just spent ages trying to figure out how to get that to show, I'm so useless at this.
Reply 3
Original post by angelooo
...


01cosπx2dx\displaystyle \int_{0}^{1}\cos \frac{\pi x}{2}dx

01cos(π2x)dx\displaystyle \int_{0}^{1}\cos (\frac{\pi}{2}x)dx

[1π2sin(π2x)]01\left[\dfrac{1}{\frac{\pi}{2}}\sin (\frac{\pi}{2}x)\right]_{0}^{1}

2π[sin(π2x)]01\dfrac{2}{\pi}\left[\sin (\frac{\pi}{2}x)\right]_{0}^{1}
(edited 10 years ago)
Reply 4
Original post by TenOfThem
01cosπx2dx\displaystyle \int_{0}^{1}\cos \frac{\pi x}{2}dx

01cos(π2x)dx\displaystyle \int_{0}^{1}\cos (\frac{\pi}{2}x)dx

[1π2sin(π2x)]01\left[\dfrac{1}{\frac{\pi}{2}}\sin (\frac{\pi}{2}x)\right]_{0}^{1}

2pi[sin(π2x)]01\dfrac{2}{pi}\left[\sin (\frac{\pi}{2}x)\right]_{0}^{1}

Thank you for you all your help, but I still can't seem to get the answer and I have no idea what I'm doing wrong. I know that the answer should be 2/π, however I just can't seem to figure out how to get there and it's driving me mad.
I'm really sorry for being such a pain, you're obviously 10x smarter than me haha.
Reply 5
Original post by angelooo
Thank you for you all your help, but I still can't seem to get the answer and I have no idea what I'm doing wrong. I know that the answer should be 2/π, however I just can't seem to figure out how to get there and it's driving me mad.
I'm really sorry for being such a pain, you're obviously 10x smarter than me haha.


What're sin(π2×1)\sin(\frac{\pi}{2} \times 1) and sin(π2×0)\sin(\frac{\pi}{2} \times 0)?
Reply 6
Original post by angelooo

but I still can't seem to get the answer and I have no idea what I'm doing wrong. I know that the answer should be 2/π, however I just can't seem to figure out how to get there and it's driving me mad.


[sin(π2x)]01=sinπ2sin0=10=1\left[\sin (\frac{\pi}{2}x)\right]_{0}^{1} = \sin \frac{\pi}{2} - \sin 0 = 1 - 0 = 1
Reply 7
Original post by joostan
What's sin(π2)×1\sin(\frac{\pi}{2}) \times 1 and sin(π2)×0\sin(\frac{\pi}{2}) \times 0

sin(π2)\sin(\frac{\pi}{2}) and 0?
Reply 8
Original post by TenOfThem
[sin(π2x)]01=sinπ2sin0=10=1\left[\sin (\frac{\pi}{2}x)\right]_{0}^{1} = \sin \frac{\pi}{2} - \sin 0 = 1 - 0 = 1

That's what I got too, but I was told that the answer should be π/2 so that threw me. I'll just go ahead with 1 and hope for the best! Thanks again for all your help, you're a star!
Reply 9
Original post by angelooo
That's what I got too, but I was told that the answer should be π/2 (sic) so that threw me. I'll just go ahead with 1 and hope for the best! Thanks again for all your help, you're a star!


Sorry

The answer is 2π\dfrac{2}{\pi}


Look back at my solution to see why
Reply 10
Original post by TenOfThem
Sorry

The answer is 2π\dfrac{2}{\pi}


Look back at my solution to see why

Sorry, but I'm lost again. Looking back at the solution I would get π2\dfrac{\pi}{2}
Can you explain further please? I know it probably seems totally obvious, but I'm just completely in the dark.
Btw I'm more than willing to pay you for this, you're basically tutoring me online at the moment so I'd be happy to pay you through paypal or something!
Original post by angelooo
Sorry, but I'm lost again. Looking back at the solution I would get π2\dfrac{\pi}{2}
Can you explain further please? I know it probably seems totally obvious, but I'm just completely in the dark.



Original post by TenOfThem
01cosπx2dx\displaystyle \int_{0}^{1}\cos \frac{\pi x}{2}dx

01cos(π2x)dx\displaystyle \int_{0}^{1}\cos (\frac{\pi}{2}x)dx

[1π2sin(π2x)]01\left[\dfrac{1}{\frac{\pi}{2}}\sin (\frac{\pi}{2}x)\right]_{0}^{1}

2π[sin(π2x)]01\dfrac{2}{\pi}\left[\sin (\frac{\pi}{2}x)\right]_{0}^{1}


2π[10]\dfrac{2}{\pi}\left[1-0\right]

2π\dfrac{2}{\pi}
Reply 12
Original post by TenOfThem
2π[10]\dfrac{2}{\pi}\left[1-0\right]

2π\dfrac{2}{\pi}

Oh my goodness, how did I not see that? I've quite possibly never felt this stupid in my entire life, hahaha. Thank you again :smile:.

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