i finally found my error. I understand the basics(for the most part) of integration.
It is the substitution and the more sophisticated equations that get me.
I cannot believe this, same problem but it is /int 4_0. I'm sorry i missed that.
To answer the question of what i am doing, first I have to figure out if this is something i can use the fund. rules of integrals or something i have to (or is preferred) substitute the function. For this one, i...
Homework Statement
2. /int 1_0 (5u^7+pi^2) dx the answer is (5/8)+pi^2
3./int 4_0 (x^(1/2))(x+1) the answer is 272/15.
Homework Equations
The Attempt at a Solution
For 2. I already have the 5/8, my question do I integrate the pi^2? I tried integrating that with no success.
For...
Homework Statement
integration 9_0 c/x^3 dx. the answer is (3/8)c.
Homework Equations
just distrib. then plug in the #'s
The Attempt at a Solution
=cx^-3
= (cx^-2)/-2=c/-2x^2
I know it has something to do with distribution.
Homework Statement
(x^2+3x-1)/x^4 the answer is,(-x^-1)-(3/2)(x^-2)+(1/3)(x^-3)+C
Homework Equations
Just reverse diferentiation
The Attempt at a Solution
=x^2+3x-1+x^-4
=(x^3)/(3)+(3x^2)/(2)-x+(1/3)(x^-3+C)
(1/3)(x^3)+(3/2)(x^2)-x+(1/3)(x^-3)+C
I know i got the answer but somehow it is...
For 1. I would have to apply l'hopital's again because it is still 0/0. So, it would be(2-2cos(2x))/(4xcos2x+2sin2x). It is still 0/0 so i have to derive one more time. It would be 4sin2x/4xcos2x+4cos2x. Apply the 0 and got 0/4 = 0. Did I do this right now?
2. a. The limit of ln(x) is -infin...
I would like to know if I did these the correct way.
Homework Statement
1.lim x->0 (1/sin 2x)-1/2x. the answer is 0.
2.lim x->0 (x^-5*ln x). The answer is -infinity.
Homework Equations
1.I used L'Hopitals theorem
2.I derived them, than L'hopitals.
The Attempt at a Solution
1.I...
Homework Statement
I search all over my textbook and did not show any (simple) examples on relation rates(seriously). Well, The first problem is, x^2+y^2=25 find dy/dt when x=3. dx/dt =4 and it says find the indicated rate and assume x>0 and y>0
Homework Equations
I did my best...
I don't know if I got this right.
I used the product rule: (x)(y(x))'+(x)'(y(x)) and for derivitive of y(x), I used the
chain rule to find the derivitive. And got(x)(1*x*x)+(x)(y(x)). -> x^3+xy(x)d/dx=0 What else am I doing wrong?
Homework Statement
I just got started on this, and am not grasping the WHOLE idea.
1.xy=25 The answer says -y/x
2.x^2+3xy+y^2=15 And this says -y^2/x^2
Homework Equations
1. dy/dx(xy)= dy/dx(25)
1=0 ???
2.dy/dx x^2+3xy+y^2= dy/dx 15
2x+3+y(dy/dx) =0...
Homework Statement
find the equation for the tangent line.
1. (e^x)(cos x) where x=0
Homework Equations
plugged 0 into the equation, (e^0)(cos 0) and got 1 for the y-coordinate. so I got the points(0,1). For the slope, I derived the equation into -(e^x)(sin x). then i plugged 0 in and got 0...
3. sinx^2/cosx^2
(cosx^2)(sinx^2)'-(cosx^2)'(sinx^2)
------------------------------------
(cosx^2)^2
(cosx^2)(2 cos x sin x)+(2 cos x sin x)(sinx^2)
----------------------------------------------
(cosx^2)^2
(2 cosx^2*cosx^3*sin x cosx^2)+(2sinx^2*cosx sinx^2*sinx^2)...
For 1, I am really not seeing how it becomes sin2t from 2sinx cosx.
For 2, I got it right, it was the derivative of e^2x that messed me up. thanks cristo.
For 3, I know ultimately the third term will be left and the other two will cancel each other out. This is what i got(by following the...
For 1. I tried the chain rule(sin x inside, square outside), and it got me 2cos(t) which is worse than my first answer.
For 2. I assumed e never changes. And I never came across a problem where e changes and I am not too sure on how to do that. Is it 2 e^1?
For 3. If I am understanding this...
Homework Statement
1. Sin^2(t) The answer states sin2t
2. (e^2x)(sin x-cos x) The answer states (e^2x)(3 sin x- cos x)
3. sec^2(x)-tan^2(x)+cos(x) The answer states -sin x
Homework Equations
1.product rule
2.product rule
3. sum and diff. rule?
The Attempt at a Solution
1. (sin...
actually that is what I was trying to say, didn't know how to write it. now I am aware the h cancels, for it to be, 5/(sqrt 5x+5h +sqrt5x). I know I cannot add (sqrt 5x+5h +sqrt5x) together because it is not like. If I am wrong about that, let me know. Am I suppose to distribute it back? I don't...
i multiplied the rad. on the numerator by (sqrt 5x + sqrt 5h) +(sqrt 5x) (since
numerator has negative sign), and i got 5x+5h-5x/h(sqrt 5x + sqrt 5h) +(sqrt
5x). both 5x's on numerator go for it to be 5h/h(sqrt 5x + sqrt 5h) +(sqrt 5x).
I'm not sure on how to get (sqrt5x)-2x from...
Homework Statement
The directions state, Use the definition to differentiate the functions given.
1)f(x)=sqrt 5x. the answer states: (square root 5x)/2x.
The directions state, a. Find the difference quotient of f. and b. f'(c) by comparing the limit of the difference quotient. Also, c is a...