- Integrate (1 + 2 sin x ) wrt x with 45 and 0 degrees as bounds8.
- Integrate 2x + 3 cos x ) wrt x with 90 and 60 degrees as bounds
These questions has bounds of degrees since trig are involved.
- Integrating term by term gives,
- integrate 1 wrt x and integrate 2sin x wrt x
- this gives x - 2cos x + c
- Now the bounds are used to get the required bounded region
- Because the curve is a trig, degrees or radians can be used
- In this case the curve has a trig part and a polynomial part so radians must be used
(1 + 2 sin x )
ReplyDelete1x+ 2 +cos x
1x-2cosx
wrtx 45 degrees
1(45) +2 cos 45
45+2(.707)
45+1.414
16.414 deg
2x + 3 cos x
ReplyDeleteintegration=
2x^2/2 +3 sin x
x^2+3sinx
between 90 and 60
x^2+3sin x
x^2+3sin90
x^2+3
Question on integration (open to anyone)
ReplyDeleteIntegrate 7 + x with respect to x
Could you also please explain how you arrived at your answer
ReplyDeleteOpen question
ReplyDeleteWhat are the two types of integrals and to which one do we add a constant of integration to our final answer ? Could someone please enlighten us ?
Open question
ReplyDeleteIf for example we asked to find the area of a curve between the points x = 2 and x = 5 which one is the upper limit and which is the lower limit and why is this so ? Someone please enlighten us ?
Open question
ReplyDeleteIntegrate (2x^2 - x^3) between x = 4 and x = 1
Could someone please explain step by step how to plot a quadratic graph . What do you do first?
ReplyDeletegreen lantern why didnt you change degrees to radians
ReplyDeleteyou only change from degrees to radian only if the ask or the question needs u to convert to readians.
ReplyDelete1. 1x+ 2 +cos x
ReplyDelete1x-2cosx
wrt x 45 degrees
1(45) +2 cos 45
45+2(.707)
45+1.414
16.414 deg
2. 2x^2/2 +3 sin x
ReplyDeletex^2+3sinx
between 90 and 60
x^2+3sin x
x^2+3sin90
x^2+3
to answer reaper..
ReplyDeletethe first question
the integral of 7+x
is
7x+x^2/2
to reaper..
ReplyDeletewhen the limits are given the upper limit is the larger number and the lower limit is the smaller of the 2 numbers..
like the upper limit is 5
and lower limit is 2
to answer reaper question..
ReplyDeletethe integration of (2x^2-x^3) with limits of 4 and 1
is
the integration is
2x^3/3-x^4/4
the upper limit is 4
and
lower limit is 1
so
[2(4)^3/3-(4)^4/4)-(2(1^3/3-(1)^4/4}
(42.67-64)-(0.66-.25)
-21.33)-(.41)
-21.74
area cannot be negative
this only means that the area found is under the x-axis.
the are is 21.74
to reaper
ReplyDeletehow to draw a quadric graph.
1. to draw a graph u must have the roots which are two point on the graph.
2. the shape could only be two shapes
3. locate the coefficient of the x^2 value.
4. if it is +ve then u will have a valley graph..with a min point. if _ve then u will have a hill with a max point.
5.to find the two point either by factorizing or if the equation cannot be factorize then use the quadratic formula.
when u have gotten the 2 point the plot then and draw the graph...
2x + 3 cos x
ReplyDeleteintegration=
2x^2/2 +3 sin x
x^2+3sinx
between 90 and 60
x^2+3sin x
x^2+3sin90
x^2+3
This comment has been removed by the author.
ReplyDelete1)
ReplyDelete(1 + 2 sin x)
1x+ 2 + cos x
1x-2cos x
wrtx 45 degrees
1(45) + 2 cos 45
45 + 2(.707)
45 + 1.414
16.414 deg
to integrate(1 +2 sin x)
ReplyDelete(1+2sinx)
1x+2+cosx
1x-2cosx
wrt(45degrees)
1(45)+2cos45
45+2(.707)
45+1.414
46.414degrees
This comment has been removed by the author.
ReplyDeleteThis comment has been removed by the author.
ReplyDeleteHi Green Lantern can u plz explain #1 & #2.
ReplyDelete1. 1x+ 2 +cos x
ReplyDelete1x-2cosx
wrt x 45 degrees
1(45) +2 cos 45
45+2(.707)
45+1.414
16.414 deg
2x + 3 cos x (integrating)
ReplyDelete2x^2/2 +3 sin x
x^2+3sinx
between 90 and 60
x^2+3sin x
x^2+3sin90
x^2+3
for ques 1
ReplyDelete1x+ 2 +cos x
1x-2cosx
wrt x 45°
1(45) +2 cos 45
45+2(.707)
45+1.414
16.414°
2x + 3 cos x
ReplyDeleteintegrating that gives=
2x^2/2 +3 sin x
x^2+3sinx
between 90 and 60
x^2+3sin x
x^2+3sin90
x^2+3
1)
ReplyDeletethe question is 1 + 2 sin x and we have to integrate this. the integral of 1 is x and the integral of 2 sin x = - 2cosx.
so the expression is now:
x - 2cosx
the bounds are 45 degrees and 0 degrees.
[(45) - 2cos45] - [(0) - 2cos(0)]
[45 - 2(0.707)] - [- 2(1)]
[45 - 1.414] - [-2]
43.6 - (-2)
43.6 + 2
45.6 degrees
2)
ReplyDelete2x + 3cosx
integrating,we get:
2x^2/2 + 3sinx
x^2 + 3sinx
the bounds given are 90 degrees and 60 degrees.
since there is a polynomial part as well as a trig part, we have to convert to radians.
1 rad = 180 degrees
therefore, 1 degree = 1/ 180 rad
therefore 90 degrees= 1/180 * 90
= 0.5 rad
therefore 60 degrees= 1/180 * 60
= 0.33 rad
x^2 + 3sinx
substituting x for the limits:
[0.5^2 + 3sin(0.5)] - [0.33^2 + 3sin(0.33)]
[0.25 + 0.0262] - [0.1089 + 0.0173]
[0.2762] - [0.1262]
0.15 sq units
in the previous post, the answer is 45.6 sq. units and not degrees as integration is used to calculate area...sorry bout that...
ReplyDeleteRE:Open question
ReplyDeleteIf for example we asked to find the area of a curve between the points x = 2 and x = 5 which one is the upper limit and which is the lower limit and why is this so ? Someone please enlighten us ?
ok reaper. remember when miss was saying that wen we integrate from a limit, it gives us the area to the left of that limit until infinity?
that is the reason we have two limits. Green Lantern was right in saying that the larger of the two values is the upper limit and the smaller is the lower limit.
however the explanation is that since you get the area to the left of the limit til infinity, if you integrate from another limit as well, you also get the area to the left of that limit until infinity.the reason we find the difference between the two areas is that this gives the area that we really want, which is the area under the curve. catch? if not u can ask for more clarification...hope this is sufficient.
2)
ReplyDelete2x^2/2 +3 sin x
x^2+3sinx
between 90 and 60
x^2+3sin x
x^2+3sin90
x^2+3
(1 + 2 sin x )
ReplyDelete1x+ 2 +cos x
1x-2cosx
with respect to x 45 degrees
1(45) +2 cos 45
45+2(.707)
45+1.414
46.414 deg
2x + 3 cos x
ReplyDeleteintegration =
2x^2/2 +3 sin x
x^2+3sinx
between 90 and 60
x^2+3sin x
x^2+3sin90
x^2+3
1 +2sinx
ReplyDelete1x+2+cosx
1x-2cosx
when it is 45 degrees
1(45)+2cos 45
46.414 degrees
1 + 2sinx
ReplyDelete1x-2cosx when 45 degrees 1(45)+2cos 45=46.414
mommy can you teach over integration tomorrow plsssssss
ReplyDelete1)
ReplyDelete(1 + 2 sin x )
1x+ 2 +cos x
1x-2cosx
wrtx 45 degrees
1(45) +2 cos 45
45+2(.707)
45+1.414
16.414 deg
2)
ReplyDelete2x^2/2 +3 sin x
x^2+3sinx
between 90 and 60
x^2+3sin x
x^2+3sin90
x^2+3
∫(1 + 2 sin x)dx
ReplyDeletex+ 2 + cos x
x-2cos x
45 degrees=Π/4
wrt to 0
1(Π/4) + 2 cosΠ/4
-1.214
wrt to 0
1(0)-2cos(0)
-2
-1.214-(-2)
=1.214 rad
help please!!!!!!
ReplyDeletei don't understand integration......can sum1 please explain.....
@ absolutely fantastic i don't understand how you arrived at your answer.....please explain....thanks
ReplyDelete1+2sinx 45,0 as bounds
ReplyDelete[x-2cosx]
[π/4-2cos(π/4)]-[0-2cos0]
π/4-1.414+2
π/4-0.586
2x + 3 cosx 90,60 as bounds
ReplyDelete2x^2 + 3sinx
[2(π/2)^2 + 3sin(π/2] - [2(π/3)^2 + 3sin(π/3)
4.935 + 3 + 2.193 + 2.598
12.726
(1 + 2 sin x )
ReplyDelete1x+ 2 +cos x
1x-2cosx
wrtx 45 degrees
1(45) +2 cos 45
45+2(.707)
45+1.414
16.414 deg
1)
ReplyDelete(1 + 2 sin x )
1x+ 2 +cos x
1x-2cosx
wrtx 45 degrees
1(45) +2 cos 45
45+2(.707)
45+1.414
16.414 deg
(1 + 2 sin x )
ReplyDelete1x+ 2 +cos x
1x-2cosx
wrtx 45 degrees
1(45) +2 cos 45
45+2(.707)
45+1.414
16.414 deg
2x + 3 cos x
ReplyDeleteintegration=
2x^2/2 +3 sin x
x^2+3sinx
between 90 and 60
x^2+3sin x
x^2+3sin90
x^2+3
(1 + 2 sin x )
ReplyDelete1x+ 2 +cos x
1x-2cosx
wrtx 45 degrees
1(45) +2 cos 45
45+2(.707)
45+1.414
16.414 deg
integration i am little bewildered about!!!
ReplyDeletei need some clarification and explanation!!!!
ReplyDelete1)
ReplyDelete1+2sinx
integration
1x+2cosx
45 degrees
1(45)+2cos45
45+1.414
46.414
question on number 2)2x+3cosx
ReplyDeletehow come u didn't subs. 90 degrees into x^2
2x^2/2+3sinx
(anyone)
number (1)
ReplyDelete1+2sinx 45,0 as bounds
[x-2cosx]
[π/4-2cos(π/4)]-[0-2cos0]
π/4-1.414+2
π/4-0.586
number (2)
ReplyDelete2x + 3 cosx 90,60 as bounds
2x^2 + 3sinx
[2(π/2)^2 + 3sin(π/2] - [2(π/3)^2 + 3sin(π/3)
4.935 + 3 + 2.193 + 2.598
12.726
1+2sinx 45,0 as bounds
ReplyDelete[x-2cosx]
[π/4-2cos(π/4)]-[0-2cos0]
π/4-1.414+2
π/4-0.586
2x + 3 cosx 90,60 as bounds
ReplyDelete2x^2 + 3sinx
[2(π/2)^2 + 3sin(π/2]-[2(π/3)^2 + 3sin(π/3)
4.935 + 3 + 2.193 + 2.598
=12.726
1). (1+2sinx)
ReplyDeletethe integral of that is x-2cosx
using bounds 45 and 0 degrees
[45-2cos45]-[0-2cos0]
[43.59]-[-2]
43.59+2
45.49
2). (2x+3cosx)
ReplyDeletethe integral of that is x^2+3sinx
using bounds 90 and 60 dgrees
[(90)^2+3sin90]-[(60)^2+3sin60]
[8100+3]-[3600+2.60]
8103-3602.6
4500.4
Integrate (1 + 2 sin x ) wrt x with 45 and 0 degrees as bounds.
ReplyDelete=1x - 2cosx
bounds 45 and 0 = 45 - 2cos 45 - 0-2cos0
= 43.59 - (-2)
= 45.59
question 1
ReplyDeletethe intergral of 1+2sinx dx
=[x-2cosx] limits 45 and 0
=[45-2cos45]-[0-2cos0]
=[43.586]-[0]
=43.586
question 2
ReplyDeletethe integral of 2x+3cosx dx
=[x^2+3sinx] limits 90 and 60
=[90^2+3sin90]-[60^2+3sin60]
=[8130]-[3602.6]
=4527.4
question 2
ReplyDeleteintegral of 1 dx
=x
integral of 2sinx dx
=-2cosx
=x-2cosx+c