# Calculus

## Questions and Answers for Calculus

Calculus

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

Calculus

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

Calculus

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

Calculus

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

Calculus

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

Calculus

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

Calculus

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

Calculus

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

Calculus

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

Calculus

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

Calculus

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

Calculus

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

Calculus

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

Calculus

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

Calculus

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

Calculus

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

Calculus

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

Calculus

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

Calculus

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

Calculus

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

Calculus

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

Calculus

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

Calculus

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

Calculus

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

Calculus

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

Calculus

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

Calculus

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

Calculus

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

Calculus

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

Calculus

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

Calculus

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

Calculus

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

Calculus

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

Calculus

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

Calculus

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

Calculus

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

Calculus

We have to find the second derivative of y = e^5x + (ln x) / 2x y = e^5x + (ln x) / 2x => y = e^5x + (1/2)(ln x)*(1/x) y' = 5*e^5x + (1/2)[(-1)(x^-2)(ln x) + (1/x^2)] y'' = 25*e^5x +...

Latest answer posted March 3, 2011 3:58 pm UTC

Calculus

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

Calculus

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

Calculus

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

Latest answer posted January 21, 2011 1:50 am UTC

Calculus

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

Calculus

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

Calculus

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

Calculus

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

Calculus

We have to prove that [(x+1)*lnx]/2 > x-1 [(x+1)*lnx]/2 > x-1 => [(x+1)*lnx] > 2(x-1) => ln x > 2(x-1)/(x+1) => ln x - 2(x-1)/(x+1) > 0 If we find the derivative of ln x -...

Latest answer posted January 21, 2011 9:47 pm UTC

Calculus

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

Calculus

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

Calculus

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

Calculus

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