Uploaded by Geron, Jose Emmanuel

INTEGRAL CALCULUS SAMPLE PROBLEMS 1

advertisement
INTEGRAL CALCULUS_ SAMPLE
PROBLEMS_ 1
NAME
erudite08
Subject
Year & Section
Gaussian integral
1. The Gaussian integral, which has useful applications in probability distributions, is given by
∫
∞
2
e−x dx =
π
−∞
Evaluate:
∫
∞
2
x2 e−x dx
0
Solution:
∞
2
I(a) = ∫0 e−ax dx
dI
da
∞
∞
2
2
= ∫0 −x2 e−ax dx = − ∫0 x2 e−ax dx
∞
2
I(a) = ∫0 e−ax dx
= 12 (
dI
da
π
)
a
∞
2
= − ∫0 x2 e−ax dx
= −(
π
2
−
= −[− 14 (
= 14 (
a = 1, 14 (
π
3
a2
π
3
12
1
3
2a 2
π
3
a2
)
)]
)
)=
π
4
Area of a circle
2. Use definite integration to show that the area enclosed by a circle of radius 1 is π .
(Hint: Consider a semi-circle above the x-axis centered at the origin.)
EQUATION OF A CIRCLE:
x2 + y2 = r 2 , since r = 1, x2 + y2 = 1 would be the equation of the circle.
x2 + y2 = 1
y = 1 − x2
AREA OF A CIRCLE (GRAPHED IN GEOGEBRA):
INTEGRAL CALCULUS_ SAMPLE PROBLEMS_ 1
1
PROVING:
1
A = 2 ∫−1
1
1 − x2 dx = 4 ∫0
1 − x2 dx , (let x = sinu, u = arcsinx, dx = cosudu)
= 4 ∫ cosu 1 − sin2 udu
= 4 ∫ cosu × cosudu
= 4 ∫ cos2 udu
For ∫
cos2 udu:
cos(u)+sinu
+ 12 ∫
2
= cos(u)+sinu
+ u2
2
∫ cos2 udu =
du
1−x2
UNDOING SUBSTITUTION: = arcsinx+x
2
1
4 ∫0
1 − x2 dx = 4( arcsinx+x
2
= 4( arcsinx+x
2
1−x2 1
∣0 )
1−x2
∣x=1 −
arcsinx+x 1−x2
∣x=0 )
2
= 4( π4 − 0)
= π → PROVED
3. Consider the region R bounded by the curves y
(mass per unit area).
= 2sin(2x) and y = 0 in the interval (0, π2 )with a uniform density k
(Graphed in GeoGebra)
a. Find the total mass of R.
b
M = pA , A = ∫a f(x) − g(x)dx
The area of the region is A
π
= ∫02 2sin(2x)dx
= −cos2x∣x= π2 − (−cos2x)∣x=0
INTEGRAL CALCULUS_ SAMPLE PROBLEMS_ 1
2
=2
Since k is the uniform density, the total mass of R is 2k .
b. Calculate the moments of inertia Mx , My of R about the x-axis and y-axis respectively.
π
Mx = ∫02 2sin2 (2x)dx
π
= ∫02 1 − cos(4x)dx
= (x − 14 sin(4x))∣x= π2 − (x − 14 sin(4x))∣x=0
= π2
π
My = ∫02 2xsin(2x)dx
π
2
= −xcos2x∣0 +
=
π
1
sin(2x)∣02
2
( Solved using integration by parts)
π
2
c. Calculate the center of mass.
x=
y=
π
2
2
π
2
2
= π/4
= π/4
The center of the mass is ( π4 , π4 )
INTEGRAL CALCULUS_ SAMPLE PROBLEMS_ 1
3
Download