Introduction to calculus:
Calculus (Latin, calculus, a small stone used for counting) 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 and integral calculus, which are related by the fundamental theorem of calculus. (Source: Wikipedia)
Example Problems for Difficult Calculus Problem
Difficult calculus example problem 1:
`Find the value of the integration`
`int_0^(5)` `int_0^(7)int_0^12`x dx`dyd``z`
Solution:
Given `int_0^(5)` x `dx``int_0^(7)int_0^12``dy``dz`
Separate the functions x, y and z, we get
= `int_0^(5)` x dx`int_0^(7)int_0^12``d`y`dz`
Integrate the coordinates separately, we get
`int_0^(5)` x dx`int_0^(7)int_0^12``d`y`dz` = `[x^2/ 2]` 05 [y]07[z]012
Substitute the limits in the above equations, in this first substitute the upper limits and then substitute lower limits.
= (`(5^2 / 2) - 0)) ` (7 - 0) (12 - 0)
= 12.5 * 2 * 5
= 125
Answer:
The final answer is 125
Difficult calculus example problem 2:
`Find the value of the integration`
`int_2^(0)` `int_0^(6)int_4^0`x dx`dyd``z`
Solution:
Given `int_2^(0)` x `dx``int_0^(6)int_4^0``dy``dz`
Separate the functions x, y and z, we get
= `int_2^(0)` x dx`int_0^(6)int_4^0``d`y`dz`
Integrate the coordinates separately, we get
`int_2^(0)` x dx`int_0^(6)int_4^0``d`y`dz` = `[x^2/ 2]` 20 [y]06[z]40
Substitute the limits in the above equations, in this first substitute the upper limits and then substitute lower limits.
= (0 - `(2^2 / 2)) ` (6 - 0) (0 - 4)
= (- 2) * 6 * (- 4)
= 48
Answer:
The final answer is 48
Difficult calculus example problem 3:
Given f (x) = 4x + 3 and g (x) be the anti derivative of f (x). Then the value of g (6) is 6 and finds the value of g (3).
Solution:
Given function f (x) = 4x + 3
Here, g (x) is anti derivative of f (x). It can be written as,
g' (x) = f (x)
Otherwise,
g (x) = `int_6^x(4x + 3)dx`
For integrating the given function, we assume h (x) is also anti derivative of f (x).
h (x) = g (x) - 6
Therefore,
h (6) = 0
we can write,
h (x) =`int_6^x(4x + 3)dx`
We find the integral value for h (3), so put x = 3
h (3) = `int_6^3(4x + 3)dx`
Integrate the above function, we get
= [`((4x^2) / (2))` + 3x]36
Substitute the limit values, we get
= [(2 * 32) + (3 * 3)] - [(2 * 62) + (3 * 6)]
= [27 - 90]
= - 63
Finally,
h (3) = g (3) - 6
rearrange the above equation, we get
g (3) = h (3) + 6
Substitute the value of h (3), we get
= - 63 + 6
g (3) = - 57
Answer:
The final answer is - 57
Please express your views of this topic 6th grade math problems with answers by commenting on blog.
Practice Problems for Difficult Calculus Problem
Practice difficult calculus problem 1:
`Find the value of the integration`
`int_0^(3)` `int_0^(4)int_0^2` dx y` dy z d``z`
Answer:
The final answer is 48
Practice difficult calculus problem 2:
Given f (x) = 7x and g (x) be the anti derivative of f (x). Then the value of g (3) is 8 and finds the value of g (7).
Answer:
The final answer is 148
Calculus (Latin, calculus, a small stone used for counting) 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 and integral calculus, which are related by the fundamental theorem of calculus. (Source: Wikipedia)
Example Problems for Difficult Calculus Problem
Difficult calculus example problem 1:
`Find the value of the integration`
`int_0^(5)` `int_0^(7)int_0^12`x dx`dyd``z`
Solution:
Given `int_0^(5)` x `dx``int_0^(7)int_0^12``dy``dz`
Separate the functions x, y and z, we get
= `int_0^(5)` x dx`int_0^(7)int_0^12``d`y`dz`
Integrate the coordinates separately, we get
`int_0^(5)` x dx`int_0^(7)int_0^12``d`y`dz` = `[x^2/ 2]` 05 [y]07[z]012
Substitute the limits in the above equations, in this first substitute the upper limits and then substitute lower limits.
= (`(5^2 / 2) - 0)) ` (7 - 0) (12 - 0)
= 12.5 * 2 * 5
= 125
Answer:
The final answer is 125
Difficult calculus example problem 2:
`Find the value of the integration`
`int_2^(0)` `int_0^(6)int_4^0`x dx`dyd``z`
Solution:
Given `int_2^(0)` x `dx``int_0^(6)int_4^0``dy``dz`
Separate the functions x, y and z, we get
= `int_2^(0)` x dx`int_0^(6)int_4^0``d`y`dz`
Integrate the coordinates separately, we get
`int_2^(0)` x dx`int_0^(6)int_4^0``d`y`dz` = `[x^2/ 2]` 20 [y]06[z]40
Substitute the limits in the above equations, in this first substitute the upper limits and then substitute lower limits.
= (0 - `(2^2 / 2)) ` (6 - 0) (0 - 4)
= (- 2) * 6 * (- 4)
= 48
Answer:
The final answer is 48
Difficult calculus example problem 3:
Given f (x) = 4x + 3 and g (x) be the anti derivative of f (x). Then the value of g (6) is 6 and finds the value of g (3).
Solution:
Given function f (x) = 4x + 3
Here, g (x) is anti derivative of f (x). It can be written as,
g' (x) = f (x)
Otherwise,
g (x) = `int_6^x(4x + 3)dx`
For integrating the given function, we assume h (x) is also anti derivative of f (x).
h (x) = g (x) - 6
Therefore,
h (6) = 0
we can write,
h (x) =`int_6^x(4x + 3)dx`
We find the integral value for h (3), so put x = 3
h (3) = `int_6^3(4x + 3)dx`
Integrate the above function, we get
= [`((4x^2) / (2))` + 3x]36
Substitute the limit values, we get
= [(2 * 32) + (3 * 3)] - [(2 * 62) + (3 * 6)]
= [27 - 90]
= - 63
Finally,
h (3) = g (3) - 6
rearrange the above equation, we get
g (3) = h (3) + 6
Substitute the value of h (3), we get
= - 63 + 6
g (3) = - 57
Answer:
The final answer is - 57
Please express your views of this topic 6th grade math problems with answers by commenting on blog.
Practice Problems for Difficult Calculus Problem
Practice difficult calculus problem 1:
`Find the value of the integration`
`int_0^(3)` `int_0^(4)int_0^2` dx y` dy z d``z`
Answer:
The final answer is 48
Practice difficult calculus problem 2:
Given f (x) = 7x and g (x) be the anti derivative of f (x). Then the value of g (3) is 8 and finds the value of g (7).
Answer:
The final answer is 148
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