Questions and Answers for Calculus

Calculus

Evaluate the integral of 1/ ( 1 + 4x^2)  

We have to find Int [1/ (1 + 4x^2) dx]. First substitute u = 2x => du /dx = 2 => du /2 = dx Now Int [1/ (1 + 4x^2) dx] => Int [(1/2)*(1/ (1+u^2) du] => (1/2)*Int [1/ (1 + u^2) du] Now...

Latest answer posted January 27, 2011 11:45 am UTC

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Calculus

What are all the critical points of f(x)=sinx+cosx ? 0=<x=<2pi

The critical points are determined by differentiating the function and equating the derivative to 0. It is solved to determine x. f(x) = sin x + cos x f'(x) = cos x - sin x = 0 => cos x = sin...

Latest answer posted February 22, 2011 2:55 am UTC

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Calculus

I need to evaluate the limit of function y=(1-cos2x)/x^2, using trigonometric identities. x approaches to 0.

You want the limit of y=(1-cos 2x)/x^2 while x approaches 0. y = (1-cos 2x)/x^2 => [1 - (1 - 2*(sin x)^2)]/x^2 => 2*(sin x)^2/x^2 => 2*(sin x / x)^2 lim x--> 0 (sin x / x) = 1 Using...

Latest answer posted April 15, 2011 2:32 am UTC

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Calculus

What is the area of the region bounded by the curve y=square root(x-1), y axis and y=1 to y=5 ?

The area of the region bounded by the curve y = sqrt (x - 1), the y- axis and the lines y = 1 and y = 5 is the limited integral of the expression of x in terms of y, between y = 5 and y = 1. y =...

Latest answer posted March 13, 2011 2:08 am UTC

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Calculus

Find the area bounded by y=8-x^2 and y=x^2.

We first determine the points where the curves y = 8 - x^2 and y = x^2, meet. 8 - x^2 = x^2 => x^2 = 4 => x = 2 , x = -2 Now we find the integral of 8 - x^2 - x^2 between the limits x = -2...

Latest answer posted February 22, 2011 2:48 am UTC

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Calculus

How to integrate cos^7 x*sinx?

We need to determine the integral of (cos x)^7 * sin x. Int [(cos x)^7 * sin x dx] let cos x = u => - du = sin x dx => Int [ -u^7 du] => -u^8 / 8 + C substitute u = cos x => - (cos...

Latest answer posted February 27, 2011 1:24 am UTC

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Calculus

Differentiate xcos2x. derivative of xcos2x is =?

We have to differentiate f(x) = x*cos 2x f'(x) = x'*cos 2x + x*(cos 2x)' f'(x) = cos 2x + x*(-sin 2x)*2 f'(x) = cos 2x - 2x*(sin 2x) The required derivative of f(x) = x*cos 2x is f'(x) = cos 2x -...

Latest answer posted May 11, 2011 6:34 am UTC

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Calculus

Evaluate the anti derivative of e^2x * cos 3x.  

We have to find Int [e^2x * cos 3x dx] Here the best way to solve would be to use integration by parts. Int [u dv] = u*v – Int [v du] take u = e^2x, du = 2*e^2x dx dv = cos 3x dx, v = (1/3)* sin...

Latest answer posted January 27, 2011 12:19 pm UTC

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Calculus

Prove that limit of the function (a^x-1)/x=lna,x->0,using two methods.  

We have to prove that lim x-->0 [(a^x - 1)/x] = ln a First, if we substitute x = 0, we get the indeterminate form 0/0. This allows the use of l"Hopital's rule and we can substitute the numerator...

Latest answer posted February 26, 2011 2:50 am UTC

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Calculus

Evaluate the limit of the function [(3+h)^2-9]/h, if h approaches to 0.

We have to find the value of lim h-->0[[(3+h)^2-9]/h] lim h-->0[[(3+h)^2-9]/h] => lim h-->0[[(3 + h - 3)(3 + h + 3)/(3 + h - 3)] cancel (3 + h - 3) => lim h-->0[[(3 + h + 3)]...

Latest answer posted March 22, 2011 6:06 am UTC

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Calculus

A curve has dy/dx=3x^2-2x. The curve passes through the point (2;5). What is the equation of the curve?

To find the curve we integrate the given dy/dx = 3x^2 - 2x. Int [ 3x^2 - 2x dx ] => 3*x^3 / 3 - 2x^2 / 2 + C => x^3 - x^2 + C As the curve passes through (2 , 5) 5 = 2^3 - 2^2 + C => 5 =...

Latest answer posted March 13, 2011 1:46 am UTC

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Calculus

What is the limit of the fraction tan4x/tan2x, x-->0 ?

We need to find the value of lim x-->0 [ tan 4x / tan 2x] If we substitute x = 0, we get the indeterminate form 0/0. This allows us to use l'Hopital's rule and substitute the numerator and the...

Latest answer posted February 20, 2011 12:37 am UTC

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Calculus

what is the limit of the function (1-sin x)/cos^2 x, x -> 90 degrees?

We have to find the value of lim x--> 90[ (1- sin x)/(cos x)^2] substituting x = 90 degrees, we get the indeterminate form 0/0, so we can use l'Hopital's rule and substitute the numerator and...

Latest answer posted February 23, 2011 2:55 am UTC

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Calculus

What is the definite integral of y=sin2x/square root(1+sin^4 x)? the upper limit of integration is pi/2 and the lower...

We have to determine the definite integral of y = sin 2x /sqrt(1 + (sin x)^4), x = 0 to x = pi/2 Int [ sin 2x /sqrt(1 + (sin x)^4) dx] let 1 + (sin x)^2 = y dy/dx = 2*sin x* cos x = sin 2x => dy...

Latest answer posted May 7, 2011 11:53 pm UTC

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Calculus

What is f'(x) if f(x)=x^(sin x)?

The function f(x) = x^(sin x) Let y = f(x) = x^(sin x) Take the natural log of both the sides ln y = ln [ x^(sin x)] => ln y = sin x * ln x Differentiate both the sides with respect to x =>...

Latest answer posted May 20, 2011 11:53 pm UTC

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Calculus

Find the equation of a curve that passes through the point (1;4) and dy/dx=4x^3+4x

We have dy/dx = 4x^3 + 4x. dy/dx = 4x^3 + 4x => dy = (4x^3 + 4x) dx Integrate both the sides Int [ dy ] = Int [ (4x^3 + 4x) dx ] => y = 4x^4 / 4 + 4*x^2 / 2 => y = x^4 + 2*x^2 + C As the...

Latest answer posted March 19, 2011 12:06 am UTC

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Calculus

Determine the limit of the function (sin5x-sin3x)/x, x-->0

We have to find the value of lim x--> 0[ (sin 5x - sin 3x)/x] if we substitute x = 0, we get the form 0/0, which allows us to use the l'Hopital's rule and substitute the numerator and the...

Latest answer posted March 3, 2011 3:36 pm UTC

1 educator answer

Calculus

What is the antiderivative of y=x/square root(x^2-9) ?

We have to find the antiderivative of y = x / sqrt ( x^2 - 9) Let u = x^2 - 9 => du / dx = 2x => x dx = du/2 Int [ x / sqrt ( x^2 - 9) dx] => Int [(1/2)*(1/ sqrt u) du] => Int [ (1/2)*...

Latest answer posted January 21, 2011 2:01 am UTC

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Calculus

Evaluate the limit of the fraction (f(x)-f(1))/(x-1), if f(x)=1+2x^5/x^2? x->1

We are given that f(x)=1+2x^5/x^2 = 1 + 2x^3. We have to find: lim x -->1 [(f(x) - f(1))/(x-1)] => lim x -->1 [(1+ 2x^3 - 1 - 2)/(x-1)] => lim x -->1 [(2x^3 - 2)/(x-1)]\ => lim x...

Latest answer posted January 22, 2011 1:43 am UTC

1 educator answer

Calculus

Evaluate the limit of the function (2x-sin2x)/x^3; x-->0.

We have to determine lim x-->0 [(2x - sin 2x)/x^3] If we substitute x = 0, we get the indeterminate form 0/0, so we use the l'Hopital's Rule and substitute the numerator and denominator with...

Latest answer posted March 14, 2011 7:04 pm UTC

1 educator answer

Calculus

What is the limit of the function sin x/(1-2sin^2x) -cos x/(2cos^2x-1) - sin2x/cosx, if x approaches to pi/4?

We need to find the value of lim x--> pi/4 [sin x/(1- 2*(sin x)^2) - cos x/2*(cos x)^2 - 1) - sin 2x / cos x. Substituting x = pi/4, gives us an indeterminate value. lim x--> pi/4 [sin x/(1-...

Latest answer posted March 24, 2011 12:16 am UTC

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Calculus

Verify if limit of ln(1+x)/x is 1, x-->0

We have to verify that lim x-->0 [ ln(1+x)/x] = 1. substituting x = 0, we get the indeterminate form 0/0, therefore we can use the l'Hopital's rule and substitute the numerator and denominator...

Latest answer posted February 25, 2011 4:07 pm UTC

1 educator answer

Calculus

Definite integral of x^2/sqrt(x^3+1), between 2 and 3, gives=?

We have to find the value of the definite integral of x^2/sqrt (x^3 + 1) between the limits x = 2 and x = 3. First we determine the indefinite integral and then substitute the values x = 3 and x =...

Latest answer posted April 4, 2011 3:44 pm UTC

1 educator answer

Calculus

How to evaluate the limit of (cos x - cos 3x) / x*sin x if x-->0 ?

We have to determine the value of lim x--> 0[(cos x - cos 3x) / x*sin x If we substitute x = 0, we get (1- 1) / 0 = 0/0 As this is an indeterminate form we use l'Hopital's rule and replace the...

Latest answer posted February 18, 2011 8:02 pm UTC

1 educator answer

Calculus

What is the area of the region enclosed by the curves y=lnx and y=ln^2x ?

First we need to determine the points of intersection between lnx and ln^2 x ==> ln x = ln^2 x ==> ln^2 x - ln x = 0 ==> lnx ( lnx -1) =0 ==> lnx = 0 ==> x = 1 ==> lnx-1 = 0...

Latest answer posted May 11, 2011 2:10 am UTC

1 educator answer

Calculus

If y = sin x + cos 3x, find the third derivative of y?

We have the function y = sin x + cos 3x. The derivative of sin x is cos x and the derivative of cos x is -sin x. Also, for a function of the form y= f(g(x)), the derivative of y or y' is given by...

Latest answer posted February 5, 2011 10:17 pm UTC

1 educator answer

Calculus

solve the limit of the function f(x)=sin5x/sinx if x --> pi

We have to find the value of lim x--> pi[ sin 5x / sin x] We see that substituting x with pi gives us the form 0/0 which is indeterminate. We can use therefore use l'Hopital's rule and use the...

Latest answer posted February 18, 2011 7:12 pm UTC

1 educator answer

Calculus

A company wishes to manufacture a box with a volume of 44 cubic feet that is open on top and is twice as long as it...

The company wishes to manufacture a box with a volume of 44 cubic feet that is open on top and is twice as long as it is wide. Let the length of the box that uses minimum amount of material be L....

Latest answer posted October 16, 2013 3:14 am UTC

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Calculus

If limit of function f(x)=(sin x-cos x)/cos 2x is l, choose the good answer: a)l=0;b)l=-1;c)=6;d)l=1/6;e)l=-square...

We have to find the limit of f(x)=(sin x-cos x)/cos 2x for x--> 45 degrees. We know that cos 2x = (cos x)^2 - (sin x )^2 lim x--> 0 [(sin x-cos x)/cos 2x] => lim x--> 0 [(sin x-cos...

Latest answer posted January 28, 2011 4:01 pm UTC

1 educator answer

Calculus

Evaluate the limit of the function ln(1+x)/(sinx+sin3x) x-->0

We have to find the value of lim x--> 0 [ ln(1+x)/(sinx+sin3x)] substituting x = 0, we get the indeterminate form 0/0. Therefore we can use l'Hopital's Rule and substitute the numerator and...

Latest answer posted February 22, 2011 1:07 am UTC

1 educator answer

Calculus

Using l'Hopital' Rule, find the limit of x^x^2 as x approaches 0 from the right.

We have to find the value of lim x-->0+ [(x^x)^2)] lim x-->0+ [(x^x)^2)] => lim x-->0+ [e^(ln ((x^x)^2))] => lim x-->0+ [e^(ln ((x^2x))] => lim x-->0+ [e^(2x*ln x)] As the...

Latest answer posted July 3, 2011 2:08 pm UTC

1 educator answer

Calculus

math Evaluate the derivative of the expression (2-x)^square root x.

We have to find y' for y = (2-x)^(sqrt x) Use natural logariths for both the sides ln y = ln[ (2-x)^(sqrt x)] use the property ln a^x = a*ln x => ln y = (sqrt x)*ln ( 2 - x) Do implicit...

Latest answer posted April 30, 2011 8:54 pm UTC

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Calculus

Using integral calculus integrate the function dy/y^2+8y+20

We have to integrate [1/ (y^2 + 8y + 20) dy] Int [1/( y^2 + 8y + 20) dy] => Int [ 1 / (y^2 + 8y + 16 + 4) dy] => Int [ 1/((y + 4)^2 + 2^2) dy] if u = y + 4 , dy = du => Int [ 1/ ( u^2 +...

Latest answer posted February 25, 2011 4:45 pm UTC

1 educator answer

Calculus

y= Arcsin^2x-2x+2√1-x^2 Arcsin X

Hi, djshan, Sorry, but I'm not too sure what you want us to do here. Are we going to graph this? Find the intercepts? Find the zeros? Something else? I would assume we are graphing it. To...

Latest answer posted March 29, 2013 5:47 pm UTC

1 educator answer

Calculus

Verify that f(x)=-2f'(x)-f"(x) if f(x)=(3x+1)*e^-x

We have f(x) = (3x + 1)*e^-x We use the product rule to find f'(x) f'(x) = (3x + 1)'*e^-x + (3x + 1)*(e^-x)' => 3*e^-x - (3x +1)e^-x => 3*e^-x - f(x) f''(x) = -3e^-x - f'(x) => -3e^-x -...

Latest answer posted February 18, 2011 8:19 pm UTC

1 educator answer

Calculus

What is the limit of the function (sin2x-sin6x)/4x, x-->0?

We have to find the value of lim x-->0 [(sin 2x - sin 6x) / 4x] We see that substituting x = 0 at this stage gives an indeterminate form 0/0. So we use l'Hopital's rule and substitute the...

Latest answer posted March 6, 2011 4:10 pm UTC

1 educator answer

Calculus

What is the second derivative of the function f(x) given by f(x)=e^2x+sin2x/2x ?

We have the function f(x) = e^2x + (sin 2x) / 2x f'(x) = [e^2x + (sin 2x) / 2x]' => (e^2x)' + [(sin 2x/2x]' => 2*e^2x + (2x*2*cos 2x - 2*sin 2x )/4*x^2 => 2*e^2x + (2x *cos 2x - sin 2x...

Latest answer posted March 14, 2011 6:02 pm UTC

1 educator answer

Calculus

Determine the indefinite integral of y=x/square root(x^2+9), using substitution.

We have to find the integral of y= x / sqrt(x^2+9) Int [ x/ sqrt (x^2 + 9) dx] let x^2 + 9 = u, du / 2 = x dx => (1/2)*Int [ 1/ sqrt u du] => (1/2)*Int [ u^(-1/2) du] => (1/2) u^(1/2) /...

Latest answer posted March 13, 2011 1:51 am UTC

1 educator answer

Calculus

Determine the gradient vector of the function f(x,y)=x^3*y^2-2x at the point (2,4)?

The gradient vector is a vector whose first component is the derivative of f(x, y) with respect to x and the second is the derivative of f(x, y) with respect to y. Here f(x, y) = x^3*y^2 - 2x...

Latest answer posted April 15, 2011 3:19 am UTC

1 educator answer

Calculus

What is the extreme value of 2x^3+3x^2-12x+5?

The extreme values of a function occur at the points where the derivative is equal to 0. f(x) = 2x^3 + 3x^2 - 12x + 5 => f'(x) = 6x^2 + 6x - 12 6x^2 + 6x - 12 = 0 => x^2 + x - 2 = 0 => x^2...

Latest answer posted February 10, 2011 5:16 pm UTC

1 educator answer

Calculus

Determine using calculus dy/dx if y =arcsin x/(1-x^2)?

We have to find the derivative of y = arc sin x/(1-x^2). We use the quotient rule here: y' = [(arc sin x)'*(1 - x^2) - ( arc sin x)*(1 - x^2)']/(1 - x^2)^2 => [sqrt(1-x^2)*(1 - x^2) + 2x*(arc...

Latest answer posted March 10, 2011 11:44 pm UTC

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Calculus

What is the extreme point of the curve 3x-6x^2?

We have to find the extreme point of the curve y = 3x - 6x^2. To do that we find the first derivative of 3x - 6x^2 and equate it to zero. This is solved for x. Now y = 3x - 6x^2 y' = 3 - 12x 3 -...

Latest answer posted January 21, 2011 1:41 am UTC

1 educator answer

Calculus

If y=(1+x^2)^3 find dy/dx.

We have y=(1+x^2)^3 We have to find dy/dx. We can use the chain rule here. dy/dx = 3(1 + x^2)^2*2x => dy/dx = 6x(1 + x^2)^2 The required result is 6x*(1 + x^2)^2

Latest answer posted January 27, 2011 3:03 am UTC

1 educator answer

Calculus

What is the area of the region enclosed between the curves y=x^2-2x+2 and -x^2+6 ?

We have to find the area enclosed between y=x^2 - 2x + 2 and y = -x^2 + 6. First lets find the points of intersection x^2 - 2x + 2 = -x^2 + 6 => 2x^2 - 2x - 4 = 0 => x^2 - x - 2 = 0 => x^2...

Latest answer posted March 2, 2011 1:29 am UTC

1 educator answer

Calculus

Using l'Hospital theorem evaluate the limit of (x^2+11x-12)/(x-1), for x->1.

We have to find the value of lim x--> 1[ (x^2+11x-12)/(x-1)]. As substituting x = 0 gives us the indeterminate form 0/0, we can use l'Hopital's theorem and substitute the numerator and the...

Latest answer posted February 26, 2011 5:15 pm UTC

1 educator answer

Calculus

What is the second derivative value for the function g(x)=sin (9x)?

We have the function g(x) = sin 9x. We have to find the second derivative of g(x). We use the chain rule here. g(x) = sin 9x => g'(x) = 9 cos 9x => g''(x) = -81 sin 9x The required second...

Latest answer posted February 26, 2011 1:40 am UTC

1 educator answer

Calculus

Prove that the functions f(x)=x+2, if x=<1 and f(x)=x^2. if x>1 are discontinuous. 

The function f(x) is defined such that f(x) = x + 2 , if x <= 1 and f(x) = x^2, if x > 1. At the point x = 1, if we approach from the left lim x--> 1- [ f(x)] = 1 + 2 = 3. If we approach...

Latest answer posted March 12, 2011 12:21 am UTC

1 educator answer

Calculus

what is dy/dx if y=(3x^4+4x^3+1)(3x^2+2x^3-1)?

We have y = (3x^4+4x^3+1)(3x^2+2x^3-1) To find the derivative dy/dx we use the product rule. y' = (3x^4+4x^3+1)'(3x^2+2x^3-1) + (3x^4+4x^3+1)(3x^2+2x^3-1)' y' = (12x^3+12x^2)'(3x^2+2x^3-1) +...

Latest answer posted February 17, 2011 5:27 pm UTC

1 educator answer

Calculus

What is the second derivative of the function f(x)=2x^3+e^2x+sin 2x-lnx?

We have the function f(x) =2x^3 + e^2x + sin 2x - ln x. We have to find f''(x). f'(x) = 6x^2 + 2e^2x + 2 cos 2x - 1/x f''(x) = 12x + 4e^2x - 4 sin 2x + 1/x^2 The required second derivative f(x) =...

Latest answer posted February 23, 2011 2:27 am UTC

1 educator answer

Calculus

Solve the derivative of expression: tan^2(ln(square root(4e^x+2x^2))) + cot^2(ln(square root(4e^x+2x^2)))

We have the expression: tan( ln (sqrt (4e^x+2x^2)))^2 + cot (ln (sqrt (4e^x + 2x^2)))^2 We have to find the derivative of the expression. We use the chain rule and start from the innermost...

Latest answer posted February 3, 2011 11:54 pm UTC

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