definite Integrals?

Question

USe geometry(not Riemann Sum) to compute the integral.

1. {14,12 xdx
2.{1,0 39 square root 1-x^2 dx
n=10, n=100, n=1000

Answer 1

EDIT: I didn't read your first sentence. But since I took the time to answer this anyway, I'm leaving this as-is.

∫(12 to 14, x dx )

To solve this, first integrate x. Then evaluate at the bounds; plug in higher bound, MINUS, plug in lower bound.

The integral of x is (1/2)x^2, so our answer is

(1/2)x^2 {evaluated from 12 to 14}

Which becomes

(1/2)14^2 - (1/2)12^2

(1/2)196 - (1/2)144

98 - 72

26

2) ∫(0 to 1, 39sqrt(1 - x^2) dx )

First, factor out the 39, as it is a constant.

39 ∫(0 to 1, sqrt(1 - x^2) dx )

Now, use trigonometric substitution.
Let x = sin(t). Then
dx = cos(t) dt.

Note that this changes our bounds too.
When x = 0, then sin(t) = 0, which means t = 0.
When x = 1, then sin(t) = 1, which means t = &pi/2.
That makes our new bounds 0 to &pi/2.

39 ∫(0 to &pi/2, sqrt(1 - sin^2(t)) cos(t) dt )

Use the identity 1 - sin^2(t) = cos^2(t)

39 ∫(0 to &pi/2, sqrt(cos^2(t)) cos(t) dt )

Taking the square root of a squared value yields itself.

39 ∫(0 to &pi/2, cos(t) cos(t) dt )

39 ∫(0 to &pi/2, cos^2(t) dt )

Now, use the half angle identity
cos^2(t) = (1/2)(1 + cos(2t))

39 ∫(0 to &pi/2, (1/2)(1 + cos(2t)) dt )

Factor the (1/2) from the integral,

(39/2) ∫(0 to &pi/2, (1 + cos(2t)) dt )

And now, this is easy to integrate.

(39/2) (t + (1/2)sin(2t)) {evaluated from 0 to &pi/2}

(39/2) [ ( &pi/2 + (1/2)sin(2*&pi/2) ) - (0 + (1/2)sin(2*0)) ]

(39/2) [ &pi/2 + (1/2)sin(&pi) - 0 ]

(39/2) [ &pi/2 + (1/2)(-1) ]

(39/2) [ &pi/2 - (1/2) ]

(39/2) [ &pi - 1 ] / 2

(39/4)(&pi - 1)

Answer 2

i'm sorry yet Gianlino's info isn't valid. x^2 and sqrt(x) are inverse of 1 yet another, although their integrals from 0 to one million are respectively one million/3 and a pair of/3 extra many times, the graphs of purposes inverse of 1 yet another would be symmetrical wrt the 1st bisector. hence the integrals, measuring the area between those graphs and the x axis and the lines x=0 and x=one million, will many times no longer be equivalent. they could be so purely in specific situations. i think of that's truly complication-free. permit me have a attempt. Dina, your substitution is virtually the the suitable option one, look With u^3 = one million-x^7 J = Int(0,one million)(one million-x^7)^(one million/3) dx transforms as follows: (one million-x^7)^(one million/3) --> u dx --> d(one million-u^3)^(one million/7) So J = Int(one million,0)ud(one million-u^3)^(one million/7) = u(one million-u^3)^(one million/7)|_1^0 - Int_1^0 (one million-u^3)^(one million/7)du = 0 + your 2nd essential so as that they are equivalent ! you need to apply the comparable trick for any (m,n) effective integers in selection to (3,7) Edited: ok, my argument approximately gianlino's replace into incorrect simply by fact i did no longer pay interest to his specifying lowering purposes. And the replace of variables I proposed is precisely replacing function and variable (hence employing the inverse function) and is an indication of gianlino's declare. i want some cafeine

Answer 3

Sorry, but what you are asking is not understandable. Wherever you got the question, please resubmit it exactly as it was written. If whhat you have presented is exactly as it was written, then get a new text book or teacher.

Answer 4

Could you put this in something resembling mathematical notation?