Introduction to AP calculus:

Calculus is a branch in mathematics focused on limits, functions, derivatives integrals and infinite series. This subject constitutes a major part of modern mathematics education. It has two major branches,

Differential calculus

Integral calculus

Both differential and integral calculus are related to the fundamental theorem of calculus.

Advanced Placement Calculus is also known as AP Calculus. It is used to indicate one of two distinct Advanced Placement courses and examinations offered by the College Board, AP Calculus AB and AP Calculus BC.

AP calculus tutoring is very interesting to the students. AP calculus tutoring covers all the topics of calculus such as differentiation and its applications, integration and its applications, limits, continuity, fundamental theorems, etc…. In tutoring, the tutors do not provides the solution to the problems instead they assist the students to do the same problem to get the result. In AP calculus tutoring, the tutors will explain each topic in step by step. Students can learn from any every in the world through online tutoring. A few example problems are given below to show how tutors solve the ap calculus problems.

Integration is an important concept in mathematics and together with differentiation, is one of the two main operations in calculus. Given a function of a real variable x and an interval [a, b] of the real line, the definite integral is defined informally to be the net signed area of the region in the xy-plane bounded by the graph of .

Examples Problems of ap calculus tutoring:

Example 1: If y = 4 – 3t2 and x = t2 – t + 1, find ‘(dy)/(dx)’ in terms of t.

Solution:

Step 1: Given

y = 4 – 3t2 x = t2 – t + 1

Step 2: Differentiate the function y = 4 – 3t2 with respect to ‘ t ‘,

‘(dy)/(dt)’ = – 6t

Step 3: Differentiate the function x = t2 – t + 1 with respect to ‘ t ‘,

‘(dx)/(dt)’ = 2t – 1

Step 4: Find ‘(dy)/(dx)’

‘ (dy)/(dx)’ = ‘(dy)/(dt)’ * ‘(dt)/(dx)’

= (- 6t) * 1/(2t – 1)

= 6t/(1- 2t)

Example 2: If y = cos x3, find ‘(dy)/(dx)’ .

Solution:

Step 1: Given

y = cos x3

Step 2: Assign variables

u = x3

y = cos u

Step 3: Differentiate the function y = cos u with respect to ‘ u ‘,

‘(dy)/(du)’ = – sin u

Step 4: Differentiate the function u = x3 with respect to ‘ x ‘,

(du)/(dx) = 3×2

Step 5: Find ‘(dy)/(dx)’

‘(dy)/(dx)’ = ‘(dy)/(du)’ * ‘(du)/(dx)’

= – sin u * 3×2

= – 3×2 sin x3

Ex 3: If f(x) = 5×6 + 2×4 + 7×2 + 8x, then find f ‘(0).

Sol : Step 1: Given function

f(x) = 5×6 + 2×4 + 7×2 + 8x

Step 2: Differentiate the given function f(x) = 5×6 + 2×4 + 7×2 + 8x with respect to ‘ x ‘,

f ‘(x) =30×5 + 8×3 + 14x + 8

Step 3: Find f ‘(0)

To find f ‘(0), plug x = 0 in f ‘(x). Therefore,

f ‘(0) = 0 + 0 + 0 + 8

= 8

Ex 4: If y = e2x – 7, then find dy/dx.

Sol : Step 1: Given function

y = e2x – 7

Step 2: Differentiate the given function y = e2x – 7 with respect to ‘ x ‘,

y = e2x – 7

‘dy/dx’ = e2x – 7 ‘d/dx’ (2x – 7) [note ‘d/dx’ (ex) = ex ]

= e2x – 7 (2)

= 2e2x -7

Example 1: Evaluate ‘int_0^(pi/3)’ sin 5x cos x dx

Solution:

Step 1: Given

‘int_0^(pi/3)’ sin 5x cos x dx

Step 2: Find the indefinite integral of sin 5x cos x

‘int’ sin 5x cos x dx = ‘1/2’ ‘int’ 2 sin 5x cos x dx

Using the formula 2 sin A cos B = sin(A + B) + sin(A – B), we can write the above function as follows

‘ int’ sin 5x cos x dx = ‘1/2’ ‘ int’ (sin 6x + sin 4x) dx

‘int’ sin 5x cos x dx = ‘1/2’ [- ‘(cos6x)/6’ – ‘(cos4x)/4’ ]

‘int’ sin 5x cos x dx = – ‘(cos6x)/12’ – ‘(cos4x)/8’

Step 3: Now, apply the limits to the integral

‘int_0^(pi/3)’ sin 5x cos x dx = [-‘(cos6x)/12’ – ‘(cos4x)/8’ ]0pi/3

= (-‘ (cos2pi)/12’ – ‘cos4pi/3/8’ ) – (- ‘1/12’ – ‘1/8’ )

= (-‘ 1/12’ + ‘1/16’ ) +(‘1/12’ + ‘1/8’ )

= ‘3/16’

Example 2: Solve ‘int’ sin 3x sin 4x dx

Solution:

Step 1: Given that the function

‘int’ sin 4x sin 3x dx

Step 2: Multiply and divide the above shown integral by 2, we get the following

‘int’ sin 4x sin 3x dx = 1/2 ‘int'( 2 sin 4x sin 3x) dx………….(1)

Step 3: We know that 2 sin a sin b = cos(a – b) – cos(a + b) ……..(2)

substitute (2) in (1).

Hence, 1/2 ‘int'( 2 sin 4x sin 3x) dx = 1/2 ‘int’ (cos x – cos 7x) dx

Step 4: By separating the integrals we get

1/2 ‘int’ (cos x – cos 7x) dx = 1/2 ‘int’ cos x dx – 1/2 ‘int’ cos 7x dx

Step 5: Integrate the above shown function with respext to ‘ x ‘ ,

1/2 ‘int’ cos x dx – 1/2 ‘int’ cos 7x dx = 1/2 sin x – (sin 7x)/14 + C

The answer is 1/2 ‘int’ cos x dx – 1/2 ‘int’ cos 7x dx = 1/2 sin x – (sin 7x)/14 + C

Example 3: Solve ‘int’ (sin3x)/(cos(3/2)x) dx

Solution:

Step 1: Given that the function

‘int’ (sin3x)/(cos(3/2x)) dx

Step 2: We know that sin 3x = 2 sin(3/2)x cos(3/2x) , then

‘int’ (sin3x)/(cos(3/2)x)dx = 2 ‘int’ sin(3/2)x cos(3/2x)/(cos(3/2)x) dx

Step 3: Cancel the cos (3/2x) term , we get the following

2 ‘int’ sin(3/2)x cos(3/2x)/(cos(3/2)x) dx= 2 ‘int’ sin(3/2)x dx

Step 4: Integrate with respect to x ,

2 ‘int’ sin(3/2)x dx = 2(- 2cos (3/2)x)/3 + C

Step 5:The answer is -‘4/3’ cos (3/2)x) + C

Example 4: Solve ‘int’ (sin3x)/(sin2 3x) dx

Solution:

Step 1: Given that the function

‘int’ (sin3x)/(sin2 3 x)dx

Step 2: divide sin 3x on both the numerator and denominator, we get

‘int’ (sin3x)/(sin2 3x) dx= ‘int’ 1/sin 3x dx

Step 3: we know that 1/sin 3x = cosec 3x

Step 4: Integrating the above we get,

‘int’cosec 3x=ln (cosec 3x + cot 3x)+c

Step 5 :So the answer is ‘int’=ln (cosec 3x + cot 3x)+c

Example 5: Find ‘int’ ‘1/x^5’ dx

Sol : Step 1: Given

‘int’ ‘1/x^5’

Step 2: Take reciprocal of ‘1/x^5’

Reciprocal of ‘1/x^5’ is x-5, so

‘int’ ‘1/x^5 ‘ = ‘int’ x-5

Step 3: Integrate the function x-5 with respect to ‘ x ‘,

Practice problems:

1) Evaluate’ int’ x’sqrt(1+x)’ dx

Ans: ‘2/5’ (1 + x)5/2 – ‘2/3’ (1 + x)3/2

2) Integrate sin’3/4′ x + 5)

Ans: – ‘4/3’ cos (‘3/4’ x + 5)

3) If y = ‘1/(a^2-x^3)^(3/2)’ , find ‘(dy)/(dx)’

Ans: ‘(9x^2)/(2(a^2-x^3)^(5/2)’

3)’int’ sin3 x dx.

Ans: ‘- cos x + cos3 x/3 + C’

4)’:”int’sin x sin 2x sin 3x dx.

Ans: – (1/16) cos4x – (1/8) cos2x + (1/24) cos6x + C