How do you evaluate the definite integral #int sinsqrtx dx# from #[0, pi^2]#?

Answer 1

#int_0^(pi^2)sin(sqrtx)dx=2pi#

Let's integrate without bounds first.

#I=intsin(sqrtx)dx#
Make the substitution #t=sqrtx#. This implies that #x=t^2#, which when differentiated shows that #dx=2tdt#. Thus:
#I=intsin(t)(2tdt)=2inttsin(t)dt#
We should then integrate this using integration by parts (IBP), which takes the form #intudv=uv-intvdu#. For the integral we're working with, we should let:
#{(u=t,=>,du=dt),(dv=sin(t)dt,=>,v=-cos(t)):}#
Recall that you differentiate #u# and integrate #dv# once you've assigned them their values.

Thus:

#I=2[uv-intvdu]=2[-tcos(t)-int(-cos(t))dt]#
#color(white)I=-2tcos(t)+2intcos(t)dt#
#color(white)I=-2tcos(t)+2sin(t)#
#color(white)I=2sin(sqrtx)-2sqrtxcos(sqrtx)#

We can now use the bounds:

#I_B=int_0^(pi^2)sin(sqrtx)dx=[2sin(sqrtx)-2sqrtxcos(sqrtx)]_0^(pi^2)#
#color(white)(I_B)=(2sin(pi)-2picos(pi))-(2sin(0)-2(0)cos(0))#
#color(white)(I_B)=(0-2pi(-1))-(0-0)#
#color(white)(I_B)=2pi#
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Answer 2

To evaluate the definite integral ( \int_{0}^{\pi^2} \sin(\sqrt{x}) , dx ), you can use a substitution method. Let ( u = \sqrt{x} ), then ( du = \frac{1}{2\sqrt{x}} dx ). Rearranging, we find ( dx = 2u , du ).

Now, when ( x = 0 ), ( u = \sqrt{0} = 0 ), and when ( x = \pi^2 ), ( u = \sqrt{\pi^2} = \pi ).

So, the integral becomes:

[ \int_{0}^{\pi} \sin(u) \cdot 2u , du ]

Now, integrate ( \sin(u) \cdot 2u ):

[ \int_{0}^{\pi} 2u \sin(u) , du ]

To evaluate this integral, you can use integration by parts:

[ u \cdot dv = 2u \sin(u) , du ] [ du = du ] [ dv = \sin(u) , du ] [ v = -\cos(u) ]

Applying integration by parts:

[ \int 2u \sin(u) , du = -2u \cos(u) - \int -2\cos(u) , du ] [ = -2u \cos(u) + 2 \sin(u) + C ]

Now, apply the limits of integration:

[ -2\pi \cos(\pi) + 2 \sin(\pi) - (-2\cdot 0 \cos(0) + 2 \sin(0)) ] [ = 2\pi ]

So, ( \int_{0}^{\pi^2} \sin(\sqrt{x}) , dx = 2\pi ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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