Calculus Questions and Answers

Calculus

The integral of ln x can be found using integration by parts. Int [ u dv ] = u*v - Int[v du] let u = ln x => du = 1/x let dv = 1 => v = x Int [ ln x * 1] = x*ln x - Int[x/x] => x*ln x - x...

Latest answer posted April 14, 2011 6:51 am UTC

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Calculus

We need to find the integral of f(x) = [1/(x(ln x)^2 + 9)] Let ln x = y => dy/dx = 1/x => dy = dx / x Int [f(x) dx] = Int [ 1/x*((ln x)^2 + 9) dx] => Int [ 1/ (y^2 + 9) dy] => [arc tan...

Latest answer posted March 23, 2011 1:27 am UTC

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Calculus

We have to find the derivative of y=(x-1)^2+2(x-1)(x+2)+(x+2)^2 Using the product rule and the chain rule: y' = 2(x - 1)*1 + 2[(x - 1)*1 + (x + 2)*1] + 2( x + 2)*1 => y' = 2x - 2 + 2x - 2 + 2x +...

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

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Calculus

To find the slant asymptote of x^3 / (x + 2)^2 we have to divide x^3 by (x + 2)^2 (x^2 + 4x + 4) | x^3...........................................| x - 4 ...........................x^3 + 4x^2 + 4x...

Latest answer posted March 23, 2011 2:15 am UTC

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Calculus

We need to find the value of lim n--> inf [ bn/2^n] Also, we have bn = 1 + 2 + 2^n lim n--> inf [ bn/2^n] => lim n--> inf [(1 + 2 + 2^n)/2^n] => lim n--> inf [1/2^n + 2/2^n +...

Latest answer posted February 22, 2011 12:08 am UTC

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Calculus

First find the derivative of f(x)=-5x^2. From first principles this is equal to lim h-->0 [(f(x + h) - f(x))/h] f(x) = -5x^2 lim h-->0 [(-5(x + h)^2 + 5x^2)/h] => -5*lim h-->0 [((x +...

Latest answer posted May 7, 2011 7:04 am UTC

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Calculus

We have to find the value of lim x-->2 [(x^3-2x^2)/(x-2)] IF we substitute x=2, we get the indeterminate form 0/0. We can use the l'Hopital's Rule and replace the numerator and denominator with...

Latest answer posted February 26, 2011 7:19 pm UTC

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Calculus

We have to find the anti-derivative of [ cos 2x - (cos x)^2]^-1 f(x) = [ cos 2x - (cos x)^2]^-1 => f(x) = [(cos x)^2 - (sin x)^2 - (cos x)^2]^-1 => f(x) = -(sin x)^-2 => f(x) = -1 / (cosec...

Latest answer posted January 24, 2011 12:01 am UTC

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Calculus

To find the slope of the tangent to the curve at the point ( pi/6 , (1/2) we have to find the first derivative of the curve and substitute the value of x at that particular point. y = sin x =>...

Latest answer posted February 1, 2011 11:19 pm UTC

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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

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Calculus

We have to find the integral of cos 2x. Int[ cos 2x dx] let 2x = u du = 2*dx dx = (1/2)*du => Int [ cos u * (1/2) du] => (1/2) sin u + C substitute u = 2x => (1/2) sin 2x + C The required...

Latest answer posted April 1, 2011 2:52 pm UTC

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Calculus

We have the function: f(x)=12x^4+24x^2+56 f(x) = 12x^4 + 24x^2 + 56 f'(x) = 48x^3 + 48x If f'(x) = 0 => 48x^3 + 48x = 0 => x^3 + 3 = 0 => x( x^2 + 1) = 0 x1 = 0 x2 = -sqrt (-1) => x2 =...

Latest answer posted January 23, 2011 1:44 am UTC

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Calculus

We have f'(x) = 25x^4 + 2e^2x f(x) = Int [ f'(x) dx] => f(x) = Int[ 25x^4 + 2e^2x dx] => f(x) = Int [ 25 x^4 dx] + Int [ 2e^2x dx] => f(x) = 25* x^5 / 5 + 2*e^2x/2 + C => f(x) = 5x^5 +...

Latest answer posted February 18, 2011 2:43 am UTC

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Calculus

You are not going to be able to write a linear equation non-linearly, because the linear part means that you are using a line. Even if you have a graph that is not a line, you can create an...

Latest answer posted January 19, 2013 11:52 pm UTC

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Calculus

The extreme value of a function is its value at the point where the value of the first derivative is equal to 0. f(x) = x^2 - 2x + 2 => f'(x) = 2x - 2 2x- 2 = 0 => x = 1 At x = 1, f(x) = 1^2...

Latest answer posted February 18, 2011 2:46 am UTC

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Calculus

To find the minimum value of x^2+x-2 we find the derivative and equate it to 0 to solve for x. Using the value of x in the function we can find the lowest value. f(x) = x^2 + x - 2 f'(x) = 2x + 1...

Latest answer posted February 11, 2011 2:21 am UTC

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Calculus

The solid that would be constructed by the curve y=(x^2) +1, the x-axis, y-axis and the line x=2 when rotated completely about x-axis, can be considered to be a series of cylinders of a height dx...

Latest answer posted February 27, 2011 12:15 am UTC

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Calculus

We have to find the integral of 1 - (sin x)^2.1- (sin x)^2 = (cos x)^2(cos x)^2 = (1/2)(1 + cos 2x)So, Int[(cos x)^2 dx]=> Int[(1/2)(1 + cos 2x) dx]=> (1/2)Int[1 dx] + (1/2)[Int(cos 2x...

Latest answer posted January 22, 2011 1:52 am UTC

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Calculus

We have the function y = e^ln(x^x) take the log of both the sides ln y = ln (e^ln(x^x)) => ln y = ln(x^x) => ln y = x*ln x differentiate both the sides (1/y)dy/dx = ln x + x/x dy/dx = y*(ln x...

Latest answer posted April 7, 2011 2:00 am UTC

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Calculus

We have the function y = 1/(x^1/3) = x^(-1/3) The integral of the function for x = 1 to x = inf. is convergent if the value is finite. Int [ x^(-1/3) dx] => x^(2/3) / (2/3) For the limits x = 1...

Latest answer posted May 1, 2011 11:27 am UTC

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Calculus

1) a) `y = 3/x` , `x>0` The area under the graph is given by `int_0^infty 3/x dx = 3log(x)|_0^infty = 3[lim_(x-> infty)log(x) - lim_(x->0)log(x)] ` `= lim_(x-> infty) x` b) `y = 12x`,...

Latest answer posted May 21, 2013 10:43 am UTC

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Calculus

We have f(x) = x^300 + x + 1 and we have to find lim x-->1 [ (f(x)-f(1))/(x-1)] => lim x-->1 [ x^300 + x + 1 - 3)/ (x-1)] for x = 1 we have the form 0/0, so we can use l'hopital's rule...

Latest answer posted February 9, 2011 2:16 am UTC

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Calculus

We need the first derivative of v(t) = ( 1 + 3^t) / 3^t v(t) = ( 1 + 3^t) / 3^t => 1/3^t + 1 => (1/3)^t + 1 v'(t) = log (1/3) *(1/3)^t The required first derivative of v(t) = ( 1 + 3^t) / 3^t...

Latest answer posted April 8, 2011 2:21 pm UTC

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Calculus

We need to find the integral of f(x) = 2x/(x+1)*(x^2+1) f(x) = 2x/(x+1)*(x^2+1) finding the partial fractions => f(x) = (Ax + B)/(x^2 + 1) + C/(x + 1) => f(x) = [Ax^2 + Ax + Bx + B + Cx^2 +...

Latest answer posted April 4, 2011 3:59 am UTC

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Calculus

To find the maximum and minimum values of the function f(x) = 2x^3 - 4x + 3, we need to find the first derivative. f(x) = 2x^3 - 4x + 3 f’(x) = 6x^2 - 4 Equate this to zero and solve for x 6x^2 - 4...

Latest answer posted January 27, 2011 11:28 am UTC

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Calculus

The function given is y = 2^(x^2 + 2x). Take the natural logarithm of both the sides ln y = ln ( 2^(x^2 + 2*x)) use the property log x^a = a*log x => ln y = (x^2 + 2*x)*ln 2 Differentiate both...

Latest answer posted April 30, 2011 12:45 pm UTC

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Calculus

The integral `int (x*e^(2x))/(2x+1)^2 dx` has to be determined. Use integration by parts. Let `x*e^(2x) = u` `du = 2*x*e^(2x) + e^(2x) dx` =>` du = e^(2x)*(1 + 2x) dx` `dx/(2x+1)^2 = dv`...

Latest answer posted August 28, 2013 1:32 pm UTC

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Calculus

We have to find lim x--> inf. [(3x^2-4x+1)/(-8x^2+5)] substituting x = inf., gives the indeterminate form inf./inf., we can use l'Hopital's rule and substitute the numerator and denominator with...

Latest answer posted May 11, 2011 2:57 pm UTC

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Calculus

We have to find the value of (x^2+2x-3)/|x-1| as x approaches from the left. As x approaches from the left x - 1 is always negative, so we have |x - 1| = (1 - x) lim x--> 1 [ (x^2+2x-3)/(1 -...

Latest answer posted March 7, 2011 9:04 pm UTC

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Calculus

Let f(x) = sqrt x f'(x) = lim h-->0 [ (sqrt (h + x) - sqrt x)/h] => lim h-->0 [ (sqrt (h + x) - sqrt x)/(h + x - x)] => lim h-->0 [ (sqrt (h + x) - sqrt x)/(sqrt (h + x))^2 - (sqrt...

Latest answer posted April 19, 2011 11:49 am UTC

1 educator answer

Calculus

For f(x)=3^4x, to find the definite integral between x = 0 and x = 1, first find the indefinite integral. F(0) = Int [ f(x) ] = 3^(4x) / 4*ln 3 The definite integral is F(1) - F(0) => 3^(4*1) /...

Latest answer posted April 19, 2011 12:02 pm UTC

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Calculus

We have the function: f(x) = (x^2-x-2)/(x-2), if x is not 2 f(x) = 1, if x=2 At x = 2, f(x) = 1 For lim x-->2 [f(x)] => lim x--> 2 [(x^2-x-2)/(x-2)] => lim x--> 2 [(x^2-2x + x -...

Latest answer posted March 22, 2011 6:24 am UTC

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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

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Calculus

We have to differentiate f(x)=(x^2-4)^(1/(x-2)) We can make the differentiation easier by eliminating the exponent on the right. For this, we take the log to the base e of both the sides. ln f(x) =...

Latest answer posted March 11, 2011 12:27 am UTC

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Calculus

x^2-4|x|+3=0 We have two options fo the absolute value. lxl = x for x > 0 lxl = -x for x < 0 Then, we have two options for the equation. x^2 - 4x + 3 = 0 OR x^2 + 4x + 3 = 0 We will solve...

Latest answer posted February 4, 2011 9:20 am UTC

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Calculus

We have to find the derivative of (x - sqrt x) / (5x^2 - 3). As we have the expression as a quotient, we use the quotient rule. f(x) = u(x) / v(x) => f'(x) = (u'(x)*v(x) - u(x)*v'(x))/v(x)^2...

Latest answer posted March 8, 2011 11:53 am UTC

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Calculus

We have the function f(x) = x^6 + 4x^4 + 3x and we have to find the second derivative. First let's find the first derivative f'(x) = 6x^5 + 16x^3 + 3 The second derivative is found by...

Latest answer posted February 22, 2011 4:52 am UTC

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Calculus

The derivative of f(x) = (cos^ 2 x)* ln (x^2) has to be found. f(x) = (cos^ 2 x)* ln (x^2) => f(x) = (cos x)^2 * ln (x^2) f'(x) = [(cos x)^2]' * ln (x^2) + (cos x)^2 * [ln (x^2)]' f'(x) = [2*cos...

Latest answer posted May 11, 2011 5:06 am UTC

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Calculus

We have to solve the equation: Int[(sin x)^2 dx]+Int[(cos x)^2 dx] = (x^3-4x)/(x-2) Now Int [ a dx ] + Int [ b dx] = Int [ (a + b) dx] Int[(sin x)^2 dx]+Int[(cos x)^2 dx] = (x^3 - 4x)/(x-2) =>...

Latest answer posted February 25, 2011 4:36 pm UTC

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