Integration by Parts MCQ Quiz in मल्याळम - Objective Question with Answer for Integration by Parts - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക
Last updated on Apr 7, 2025
Latest Integration by Parts MCQ Objective Questions
Top Integration by Parts MCQ Objective Questions
Integration by Parts Question 1:
Evaluate:
Answer (Detailed Solution Below)
Integration by Parts Question 1 Detailed Solution
Concept:
Calculus:
Derivatives of Trigonometric Functions:
Integration by Parts:
∫ f(x) g(x) dx = f(x) ∫ g(x) dx - ∫ [f'(x) ∫ g(x) dx] dx.
Calculation:
Let I =
Considering x as the first function and sin x as the second function, we get:
⇒ I = x ∫ sin x dx - ∫ [(1) ∫ sin dx] dx
⇒ I = x (-cos x) - ∫ (-cos x) dx
⇒ I = sin x - x cos x + C.
Integration by Parts Question 2:
∫ ex cos x dx =
Answer (Detailed Solution Below)
Integration by Parts Question 2 Detailed Solution
Concept:
Integration by parts: Integration by parts is a method to find integrals of products.
The formula for integrating by parts is given by:
⇒
where u is the function u(x) and v is the function v(x)
ILATE rule is Usually, the preference order of this rule is based on some functions such as Inverse, Logarithm, Algebraic, Trigonometric and Exponent.
Calculation:
I = ∫ ex cos x dx
Integrating by ILATE rule,
I = cos x ∫ ex dx - ∫ (-sin x ∫ ex dx) dx
I = ex cos x + ∫ ex sin x dx
I = ex cos x + sin x ∫ ex dx - ∫ (cos x ∫ ex dx) dx
I = ex cos x + ex sin x - ∫ ex cos x dx
I = ex cos x + ex sin x - I
2I = ex cos x + ex sin x
I =
Integration by Parts Question 3:
Evaluate:
Answer (Detailed Solution Below)
Integration by Parts Question 3 Detailed Solution
Concept:
where C is a constant
Integration by parts:
The formula for integrating by parts is given by;
Where u is the function u(x) and v is the function v(x)
- ILATE Rule: Usually, the preference order of this rule is based on some functions such as Inverse, Logarithm, Algebraic, Trigonometric and Exponent.
Calculation:
Here, we have to find the value of
According to the integration by parts (ILATE Rule) we have: x as our first function and e3x as our second function
i.e u(x) = x and v(x) = e3x
As we know that,
As we know that,
As we know that,
Integration by Parts Question 4:
The value of the integral
Answer (Detailed Solution Below)
Integration by Parts Question 4 Detailed Solution
Concept:
Greatest Integer Function [X]: It indicates an integral part of the real number x which is the nearest and smaller integer to x. It is also known as the floor of X.
[x] = The largest integer that is less than or equal to x.
X | [X] |
0 ≤ x | 0 |
0 ≤ x | 1 |
0 ≤ x |
2 |
Calculation:
⇒ I = 2 - √2
Hence, the value of the integral is 2 - √2.
Integration by Parts Question 5:
Solve ∫ (x ln x) dx
Answer (Detailed Solution Below)
Integration by Parts Question 5 Detailed Solution
Concept:
Integral property:
- ∫ xn dx =
+ C ; n ≠ -1 + C - ∫ ex dx = ex+ C
- ∫ ax dx = (ax/ln a) + C ; a > 0, a ≠ 1
- ∫ sin x dx = - cos x + C
- ∫ cos x dx = sin x + C
Integration by parts: Integration by parts is a method to find integrals of products. The formula for integrating by parts is given by:
⇒
where u is the function u(x) and v is the function v(x)
ILATE rule is Usually, the preference order of this rule is based on some functions such as Inverse, Logarithm, Algebraic, Trigonometric and Exponent.
Calculation:
I = ∫ (x ln x) dx
Using integration by parts
⇒ I = ln x ∫ x dx - ∫
⇒ I =
⇒ I =
⇒ I =
Integration by Parts Question 6:
equals
Answer (Detailed Solution Below)
Integration by Parts Question 6 Detailed Solution
Given:
The given integral is
Formula Used:
Calculation:
We have,
⇒ I =
Now |1 - x| = 1 - x for 1 - x ≥ 0 ⇒1 ≥ x and
|1 - x| = - (1 - x) for 1 - x
⇒ I =
⇒ I =
⇒ I =
⇒ I =
⇒ I =
⇒ I =
⇒ I =
⇒ I =
∴
Integration by Parts Question 7:
If
Answer (Detailed Solution Below)
Integration by Parts Question 7 Detailed Solution
Explanation:
Given Integral is,
Considering
Hence by Integration by parts we know that,
∫ [I1][ I2 ] dt = I1∫ I2 dt - ∫ [I'1∫ I2 dt]dt
Integration by Parts Question 8:
What is
Answer (Detailed Solution Below)
Integration by Parts Question 8 Detailed Solution
Concept:
Integration by parts
Integration by parts is used to integrate the product of two or more functions. The two functions
to be integrated f(x) and g(x) are of the form
integration. Among the two functions, the first function f(x) is selected such that its derivative
formula exists, and the second function g(x) is chosen such that an integral of such a function exists.
A useful rule of integral by parts is ILATE.
I: Inverse trigonometric functions : sin-1(x), cos-1(x), tan-1(x)
L: Logarithmic functions : ln(x), log(x)
A: Algebraic functions : x2, x3
T: Trigonometric functions : sin(x), cos(x), tan (x)
E: Exponential functions : ex, 3x
Properties of logarithms
Let y =
Taking log both sides,
⇒ log y = log
⇒ log y = log x log e
⇒ log y = log x
⇒ y = x
Calculation:
I =
⇒ I =
Taking x as the first function and sin x as the second function and integrating by parts, we get,
⇒ I = x
⇒ I = x (− cos x) −
⇒ I = −x cos x + sin x + C
∴ The value of
Integration by Parts Question 9:
What is ∫ (elog x + sin x) cos x dx equal to?
Answer (Detailed Solution Below)
Integration by Parts Question 9 Detailed Solution
Concept:
1. Integration by parts:
Integration by parts is a method to find integrals of products
The formula for integrating by parts is given by;
Where u is the function u(x) and v is the function v(x)
2. ILATE Rule:
Usually, the preference order of this rule is based on some functions such as Inverse, Logarithm, Algebraic, Trigonometric and Exponent.
Calculation:
Let I = ∫ (elog x + sin x) cos x dx
⇒ I = I1 + I2 .... (1)
Now, I1 =
Applying by parts, we get
⇒ x sin x -
⇒ x sin x + cos x + c
Now, I2 =
Let sin x = t
Differentiating with respect to x, we get
⇒ cos x dx = dt
⇒ I2 =
⇒
Put the value of I1 and I2 in equation (1), we get
∴ The required integral (I) is
Integration by Parts Question 10:
Find the value of
Answer (Detailed Solution Below)
Integration by Parts Question 10 Detailed Solution
Concept:
where C is a constant
Calculation:
Here we have to find the value of
Now the given integral can be re-written as:
Now by comparing the above equation with
As we know that,