Definition of Calculus:

Calculus is the rates of change, measurement of changing quantities. Calculus created Newton and Leibniz. Calculus are two types: 1) Differentiation 2) Integration. Now we see the calculus test answers exam. Some specific purposes of calculus are finding the slope of curves, calculating the area or volume of complex figures, visualizing graphs, finding the average value of a function, etc.

## Calculus test answers exam : Calculus Test Formulas:

Some Important Differentian and Integration Formulas:

Differentiation formulas

1. (`d/dx` )(xn)=nxn-1

2. (`d/dx` )(sin x)=cos x

3. (`d/dx` )(cos x)=sin x

4. (`d/dx` )(tan x)=sec2 x

5. (`d/dx` )(ex)=ex

6. (`d/dx` )(ax)=ax ln a

7. (`d/dx` )(cot x)=-csc2 x

8. (`d/dx` )(sec x)=sec x tan x

9. (`d/dx` )(csc x)=-csc x cot x

Integration formulas

1. a dx=ax+C

2. ∫ex dx=ex +C

3. sin x dx=-cos x+C

4. cos x dx=sin x+C

5. tan x dx=ln|sec x|+C

6. cot x dx=ln|sin x|+C

7. sec x dx=ln|sec x+tan x|+C

8. csc x dx=ln|csc x-cot x|+C

## Calculus test answers exam: Calculus Test Example Problems:

Calculus Test Example 1:

Differentiate y = sin(2x) + cos2x.

To eleminate using the chain rule, recall the trigonometry identity sin(2x) = 2sinx cosx, and first rewrite the problem as

y = sin(2x) + cos2x

= 2sinx cosx + cosx . cosx

Now apply the product rule twice. Then

yI = (2sinx D{cosx} + D{2sinx}cosx) + (cosx D{cosx} + D{cosx}cosx)

= 2sinx(-sinx) + (2cosx)cosx + cosx(-sinx) + (-sinx)cosx

= -sin2x + 2cos2x – 2sinx cosx

(This is an acceptable answer. By using the trigonometry identity cos2x – sin2x = cos(2x) an alternative answer can be gotten)

= 2(cos2x – sin2x) – (2sinx cosx)

= 2(cos(2x)) – (sin(2x))

= 2cos(2x) –sin(2x)

Calculus Test Example 2:

Integrate ∫ (3 –x)10 dx

Let u = 3-x

So that du = (-1) dx  or  (-1) du = dx .

Replacing all forms of x, getting

∫ (3-x)10 dx = ∫ u10(-1) du

= - ∫ u10 du

= (-u11/11) + C

= (-1/11) (3 – x)11 + C

Calculus Test Example 3:

Integrate ∫(2x + 5) (x2 + 5x) 7dx

Let u = x2+5x

so that du = (2x+5)dx

replacing all forms of x, getting

∫(2x + 5) (x2 + 5x) 7dx = ∫(x2 + 5x) 7(2x + 5)dx

= ∫u7du

= (u8/8) + C

= (1/8) (x2 + 5x)8 + C

## Calculus Practice Test Exam:

Test Exam Problem 1: Integrate ∫ `(3)/(4 + x 1 / 3)` dx

Answer:  (9/2)x2/3 – 36x1/3 + 144 ln |x1/3 + 4| + C

Test Exam Problem 2: y = `(1 + arctan x)/(2-3 arctan x)`
Answer: `(5)/((1 + x2) (2-3 arctanx)2)`