Description

This mock test of Definite Integrals MCQ - 1 for Mathematics helps you for every Mathematics entrance exam.
This contains 10 Multiple Choice Questions for Mathematics Definite Integrals MCQ - 1 (mcq) to study with solutions a complete question bank.
The solved questions answers in this Definite Integrals MCQ - 1 quiz give you a good mix of easy questions and tough questions. Mathematics
students definitely take this Definite Integrals MCQ - 1 exercise for a better result in the exam. You can find other Definite Integrals MCQ - 1 extra questions,
long questions & short questions for Mathematics on EduRev as well by searching above.

QUESTION: 1

Let f: be defined as f (t) = f(t )dt

Solution:

QUESTION: 2

A minimum value of is

Solution:

**Correct Answer :- d**

**Explanation** : f(x) = ∫(0 to x) te^{-t^2 }dt

f'(x) = xe^{-x^2} = 0

Therefore, x = 0

f"(x) = e^{-x^2}(1 - 2x^{2})

f"(0) = 1 > 0

Therefore, the minimum value is f(0) = 0

QUESTION: 3

Let f(x) = ∫e^{x}(x−1)(x−2)dx then f decrease in the interval

Solution:

f(x) = ∫e^{x}(x−1)(x−2)dx

For decreasing function ,f′(x)<0,f′(x)<0

⇒e^{x}(x−1)(x−2) < 0

⇒(x−1)(x−2) < 0

⇒ e^{x}(x−1)(x−2) < 0

⇒ (x−1)(x−2) < 0

⇒1 < x < 2,

∵ e^{x }> 0 ∀ x ∈ R

QUESTION: 4

The value of x_{1} and x_{2} with x_{1} < x_{2} such that has the largest value are

Solution:

QUESTION: 5

Let f(x) = Then the real roots of tlie equation x^{2} — f'(x) = 0 are

Solution:

QUESTION: 6

is equal to

Solution:

**ANSWER :- A**

**Solution :- limx→0 1/x[∫y→a e^sin2tdt− ∫x+y→a e^sin2tdt]**

** limx→0 [∫y→a e^sin2tdt− ∫x+y→a e^sin2tdt]/x**

**Hence it is 0/0 form, Apply L-hospital rule.**

**lim x-->0 0 - {e^(sin^2a).0 - e^(sin^2(x+y).1}/1**

**lim x-->0 e^(sin^2(x+y))**

**= e^(sin^2y)**

QUESTION: 7

If f(x) is differentiable and equals

Solution:

QUESTION: 8

If

Solution:

QUESTION: 9

The area of the region between the curves y = and bounded by the lines x = 0 and

Solution:

QUESTION: 10

Let f(x) = where g is a real valued continuous function on. Then f′ (x) is equal to

Solution:

Here f(x) = where g is a real valued continuous function on R.

Apply Lebnitz's rule, we obtain

f′ (x) = 0

### Definite Integrals - 1

Video | 21:50 min

### Definite integrals

Doc | 1 Page

### Computing Definite Integrals

Doc | 22 Pages

### Definite Integrals - 3

Video | 10:08 min

- Definite Integrals MCQ - 1
Test | 10 questions | 30 min

- Definite Integrals MCQ - 2
Test | 20 questions | 60 min

- Test: Properties Of Definite Integrals
Test | 5 questions | 10 min

- Test: Evaluating Definite Integrals
Test | 10 questions | 10 min

- Test: Problems On Definite Integrals
Test | 10 questions | 10 min