Using the Riemann sum formula: A = ∫ [f(x) - g(x)]dx from a to b Find the area between y=1/2sec²t and y= -4sin²t between the points π∕3 and - π∕3 Please show a detailed solution. Thank you.
Using the Riemann sum formula: A = ∫ [f(x) - g(x)]dx from a to b Find the area between y=e^x/2 and y=e^x Between the x values of 2ln2 and zero.
Please see the attached file for the full problem description. --- 1. Transform the given integral in Cartesian coordinates to one in polar coordinates and evaluate the polar integral. : refer to integral 5. 2. Determine the values of the given integrals, where W is the region bounded by the two spheres x^2 + y^2 + z^2 = a^
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Assuming r, θ are the polar coordinates, change the order of integration: ∫-pi/2-->pi/2 ∫0-->a cos θ f(r, θ ) dr dθ Find the volume of the ellipsoid: x^2/a^2 + y^2/b^2 + z^2/c^2 ≤ 1 Let a and b be any numbers such that a^2 + b^2 =1 and f(x,y) be a continuous function of one variable. Perform the change of
You need to choose between making a public offering and arranging a private placement. In each case the issue involves $10 million face value of 10-year debt. You have the following data for each: A public issue: The interest rate on the debt would be 8.5 percent, and the debt would be issued at face value. The underwriti
∫(pi/2 to 0) sin^4(x) dx -- do not use reduction forumulas use 1-cos2u/2=sin^2x ∫3x^3/sqrt(8-x^2) dx Integrate, integration
Evaluate each of the following integrals: 1. ∫0-->2 6/(5x+2) dx 2. ∫1-->3 e^(-0.4t) dt 3..... 4.... Please see the attached file for the fully formatted problems. Integrate, Integration
Problem 1 and 2: Sketch the region of integration, reverse the order of integration, and evaluate both iterated integrals. ∫0-->2 ∫0-->4-y2 x dx dy ∫0-->pi/2 ∫0-->cosx sin x dy dx Problem 3: When you reverse the order of integration, you should obtain a sum of iterated integrals. Make the reversals and
F(x,y,z)=y ; W is the region bounded by the plane x+y+z=2, the cylinder x^2 + z^2 = 1, and y=0. Integrate the given function over the indicated region W.
1. Using the integral ∫-1-->1 ∫x^2-->1 ∫0-->1-y dz dy dx a) Sketch the region of integration. Write the integral as an equivalent iterated integral in the order: b) dy dz dx c) dx dz dy d) dz dx dy 2. Find the volume of a wedge cut from the cylinder x^2 +y^2 =1 by planes z=-y and z=0. Please show me t
I have a function (see attached). I need to integrate it over m from - infinity to infinity, h from - infinity to infinity. I need to apply a technique such that the integral takes a simple form, easy for integration. The main problem here as you can see is product of terms in the denominator. See the attached file.
7.5 Inverse trigonometric functions Find the exact value of the expression. 1) sin^-1 (SQRT3 / 2) 2) arctan(-1) 3) tan^-1 (SQRT 3) 4) cos^-1 (-1) 5) csc^-1 (2) 6) arcsin(-1/ (SQRT 2) 7) sec^-1 (SQRT 2) 8) arccos(cos 2pi) 9) tan^-1 (tan 3pi/4) 10) cos(arcsin ½) 11) sin(2 tan^-1 SQRT 2) 12) cos(tan^-1 (2) + tan
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Evaluate the integral 1) ∫ (sin^3 (x)) (cos^2 (x)) dx 2) ∫ ( sin^4 (x)) (cos^5 (x)) dx 3) ∫ ( sin^6 (x)) (cos^3 (x)) dx 4) ∫ ( sin^3 (mx)) dx 5) ∫ (from 0 to pi/2 on top) (cos^2 (theta)) dtheta 6) ∫ (from 0 to pi/2 on top) (sin^2 (2theta)) dtheta 7) ∫ (from 0 to pi on top) (sin^4 (3t)) dt 8) ∫ (from 0
Evaluate the integral using integration by parts with the indicated choices of u and du. 1) ∫ x ln x dx, u=ln x, du=xdx 2) ∫ theta sec^2(theta) dtheta, u=theta, du=sec^2(theta) dtheta Evaluate the integral 1) ∫ x cos 5x dx 2) ∫ (x)(e)^(-x) dx 3) ∫ re^(r/2) dr 4) ∫ t sin 2t dt 5
The surface S is that part of the spheroid .... which lies inside the paraboloid az = x2 + y2; here a is a constant. Sketch the surface S and the paraboloid by drawing their intersections with the plane y = 0. Show that Z Z ..... where R is a region of the (x; y) plane you should find, and hence evaluate the surface integra
Please and explain and solve the following: 13. Find the indefinite integral and check the result by differentiation. Integral: x^2(x^3 - 1)^4 dx Answer: (x^3 - 1)^5/15 + C 130. Find the indefinite integral in two ways. Explain any difference in the forms of the answers. Integral: sin x cos x dx
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I am confused as to how to solve this. Please show all steps. f(x)=1/2sec²t g(x)=-4sin²t limits: a=π/3, b=-π/3
I am not sure about the solution. Please show all steps. f(x)=e^x/2 g(x)=e^-x/2 limits: a=0, b=2ln2
I am not sure how to solve this. Please show all steps. F(x)=ln2x G(x)=lnx Limits: a=1 and b=5
Please show all steps to solve: domain: 1/2≤t≤1 ∫6dt∕√(3+4t-4t²)
Please show all steps to solve: Domain: 1≤t≤ e^π∕4 ∫4dt∕t(1+ln²t)
Please show all steps to solve: integral square root tan^-1 x dx/1 + x^2
Please show all steps to solve: ∫dx∕(x-2)√(x²-4x+3)
The question asks if both of these integrations can be correct and why/why not? a) ∫dx / √1 - x² = -∫-dx/ √1-x² = -cos‾¹x + C b) ∫dx / √1 - x² = ∫-du/√(1 - (-u)²) x = -u dx= -du = ∫-du/√1- u² = cos‾¹u + C u= -x = cos‾¹(-x) + C
What is the solution? please show each step ∫ dy / (sin‾¹y)√(1 - y²)
How is this solved? Please show the steps ∫√tan-¹x dx / 1 + x²
How do you solve this integral in the domain shown? Please show each step. domain: ½ ≤ t ≤ 1 ∫ 6dt ∕ √(3 + 4t - 4t²)