Math Integral Homework Help

Math Homework Help?!(integrals)?

You're using six rectangles, so divide the base (x-axis) by 6 to find the width of each rectangle.
(48 - 0) / 6 = 8

Each rectangle will have a base of 8 and height of f(x). The height will be calculated 3 different ways;

"L6" means the height of the rectangle will be calculated from the six left end ponts which are x = 0, 8, 16, 24, 32, and 40. So the height of the 6 rectangles will be f(0), f(8), f(16), f(24), f(32), and f(40). The area will be the base x height for each rectangle. Then just add up the 6 areas to get the estimated total area under the graph.

"R6" means the height of each rectangle will be calculated from the six right end points which are x = 8, 16, 24, 32, 40, and 48.

"M6" means use the six midpoints for height which are x = 4, 12, 20, 36, and 44.

Maths - integration by parts homework help?

∫ e^(4x) (2x+1) dx
= 2 ∫ x e^(4x) dx + ∫ e^(4x) dx

2 ∫ x e^(4x) dx :
Let t= 4x
dt = 4 dx
dx = (1/4) dt
x= t/4

2 ∫ x e^(4x) dx = (2/4) (1/4) ∫ t e^t dt
= (1/8) ∫ t e^t dt

Integrate by parts
dv= e^t dt; v= e^t
u= (1/8) t ; du = (1/8) dt

∫ u dv = u v - ∫ v du
(1/8) ∫ t e^t dt = (1/8) t e^t - (1/8) ∫ e^t dt
(1/8) ∫ t e^t dt = (1/8) t e^t - (1/8) e^t

replace t by 4x
= (1/8) (4x) e^(4x) - (1/8) e^(4x)
= (1/2) x e^(4x) - (1/8) e^(4x)

2 ∫ x e^(4x) dx = (1/2) x e^(4x) - (1/8) e^(4x) ----------- (1)

∫ e^(4x) dx
Let t= 4x
dt = 4 dx
dx = dt/4

∫ e^(4x) dx = (1/4) ∫ e^t dt
= (1/4) e^t
= (1/4) e^(4x) ------------ (2)

∫ e^(4x) (2x+1) dx = (1) + (2)

= (1/2) x e^(4x) - (1/8) e^(4x) + (1/4) e^(4x)
= (1/2) x e^(4x) + (1/8) e^(4x) + C

1c)
∫ (1+ln x) / x dx

Let u= 1+ln x
du = 1/x dx

∫ (1+ln x) / x dx = ∫ u du
= (1/2) u^2
= (1/2) (1+ ln x)^2 + C

Math homework help please?

y=sin(6x)
the curve meets x axis at y=0
thus x=0,pi/6...
now integrating y=sin(6x) with limits x=0 to pi/6
hence we have
area=-cos(6x)/6
applying limits we have
area=(-cos(pi))/6+(cos(0))/6
area= -(-1)/6+1/6
=2/6=1/3
thus area = 1/3 square units
happy to help!!happy new year!!

Math homework help for velocity?

1. The velocity of the flow of blood at a distance r from the central axis of an artery of radius r is v=k(R^2-r^2)
where k is the constant of proportionality. Find the average rate of flow of blood along a radius of the artery. (Use 0 and r as the limits of integration).

2. A company introduces a new product, and the profit in thousands of dollars over the first 6 months is approximated by the model

p=5(sqrt. of t +30, where t=1,2,3,4,5,6

(a) Complete a table and use it to calculate (arithmetically) the average profit over the first 6 months.

(b) Find the average value of the profit function by integration and compare the result with the result in part (a). (Integrate over the interval [0.5,6.5])

(c) What, if any, is the advantage of using the approximation of the average given by the definite integral? (Note that the integral approximation utilizes all real values of in the interval rather than just the integers.)

MATH HOMEWORK HELP!!!! DATA MANAGEMENT?

semi-perimeter = 18m
the 2 sides (X , Y) & the area can take the following values
X ..... Y ..... A
1 ..... 17 .... 17
2 ..... 16 .... 32
3 ..... 15 .... 45
4 ..... 14 .... 56
5 ..... 13 .... 65
6 ..... 12 .... 72
7 ..... 11 .... 77
8 ..... 10 .... 80
sub-total ... 444
9 ...... 9 ..... 81

X>9 will result in a repeat of areas obtained with X<9
since the orientation of the rectangle isn't mentioned, X can take any integer value 1-18
also, remember that a square, too, is a rectangle (with equal sides)

total = 2*444 + 81 = 969 m^2 in 17 trials
expected area = 969/17
= 57 m^2
-------------

Calculus homework help?

∫ t^3 e^t dt

u = t^3
du = 3t^2 dt
dv = e^t dt
v = e^t

t^3e^t - ∫ e^t 3t^2 dt
t^3e^t - 3(t^2e^t - ∫ e^t 2t dt)
t^3e^t - 3t^2e^t + 6 ∫ te^t dt
t^3e^t - 3t^2 e^t + 6(te^t - ∫ e^t dt)
t^3e^t - 3t^2e^t + 6te^t - 6e^t + C

∫ 6tsin(19t) dt [0, pi]

u = t
du = dt
dv = sin(19t) dt
v = (-1/19)cos(19t)

6[(-t/19)cos(19t) - ∫ (-1/19)cos(19t) dt]
6[(-t/19)cos(19t) + (1/19) ∫ cos(19t) dt]
6[(-t/19)cos(19t) + 1/(19)(19) sin(19t)]

sin(pi) = 0.

f(x1) - f(x2) = (-6/19)(pi*cos(19pi) - (-6/19)(pi*cos(19pi))
(-6/19)(pi*cos(19pi) + (6/19)(pi*cos(0))
(-6/19)(pi*cos(19pi)) + (6/19)

Math Homework Due Very soon!! Evaluate the surface integral?

Evaluate the surface integral

S F · dS
for the given vector field F and the oriented surface S. In other words, find the flux of F across S
F(x, y, z) = xy i + 8x^2 j + yz k

S is the surface
z = xe^y, 0 ≤ x ≤ 1, 0 ≤ y ≤ 3,
with upward orientation