Calculus Test Prep

calculus test prep

Introduction to calculus test preparation :

                 The common definition of Calculus is study of limits, derivatives, integrals, and infinite series. Limits are simple mathematical tool in calculus. Calculus can be used to solve differentiation of any arbitrary equation and the corresponding output result. In calculus, Let y= f(x) be a given continuous function. Then, y depends upon the x value and it changes with a change in the value of x. We can use the word increment to denote a small change of x and y.

 

Calculus formulas - calculus test prep:

 

Integral calculus formula:

1. `int` x n dx = `(x^n+1) / (n+1) ` + c

2. `int` cos x dx = sin x + c

3. `int`sin x dx = - cos x + c

4.  `int` sec2x dx = tan x + c

5..` int` cosec x . cot x dx = -cosec x + c

6. `int` sec x.tan x dx = sec x + c

7. `int ` cosec2 x dx = -cot x + c

8.  `int`` (1/x)` dx = log x + c

9.  `int` e x dx = e x + c

10. `int` u dv = uv -`int` v du

Differential calculus formula:

1.  `d / dx` (x n ) = n xn-1

2. ` d/dx` (sin x) = cos x

3.  `d/dx` (cos x) = −sin x

 4.  `d/dx` (tan x) = sec2x

5.  `d/dx` (cosec x) = −cosec x . cot x

6.  `d/dx` (sec x) = sec x . tan x

 

calculus test preparation problems :

 

Problem 1 on calculus test prep:

Integrate: `int` (-2x2 + x3 + 2) dx

Solution:

         Given `int`(-2x2 + x3 +2) dx.

             `int`(-2x2 + x3 +2) dx. =`int`(-2)x2 dx +`int`x3dx + `int`2 dx.

                                        = (-2)`int` x2 dx + `int` x3 dx + 2`int`dx.

                                        = (-2) (x3/3) + (x4/4) + 2 (x) + c.

                                        = (-2/3) x3 + (1/4) x4 + 2 x + c.

          `int` (-2x2 + x3 +2) dx = (-2/3) x3 + (1/4) x4 + 2 x + c.

       Answer:

         `int` (-2x2 + x3 +2) dx = (-2/3) x3 + (1/4) x4 + 2 x + c.

 

Problem 2 on calculus test prep:

       Differentiate with respect to x :  y =  3x5 + x3 - 2x

   Solution:

           Given, y = 3x5 + x3 - 2x

                          ` (dy)/dx` = `d/dx` (3x5 + x3 - 2x)

                                    = (5 × 3) x(5-1)  + 3x(3-1) - (2 × 1) x0 .

                                    = 15 x4 + 3x2 -2 

      ` d/dx ` (3x5 + x3 - 2x) = 15 x4 + 3x2 -2 .

     Answer: 

     ` d/dx ` (3x5 + x3 - 2x)  =  15 x4 + 3x2 -2

Practice problems on calculus test prep:

Calculus test prep practice problem 1:  

    Integrate: `int` cos 5x. cos 3x dx

     Answer: `(1/2)[((sin 8x )/8) + ((sin 2x)/2) ] ` + c

Calculus test prep practice problem 2:

Differentiate the given equation with respect to t. y = 10 t5 - 4t3 + 3t.

    Answer:

         ` d/dx` (10 t5 - 4t3 + 3t) = 50t4 -12t2 + 3

Problem 3 on calculus test prep:   

        Integrate: `int` sin 3x cos 2x .dx

     Answer: ` - 1/10` ( cos 5x + cos x) + c