Mathematics Crash Course JEE Advanced>Chapter 23 - Definite Integration>Exercise 1>Q 44

HARD

Mathematics

IMPORTANT

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Let $f (x)$ be an odd continuous function which is periodic with period $2$ . If $g (x) = \int_{0}^{x} f (t) d t$ , then

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Important Questions on Definite Integration

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Mathematics

IMPORTANT

Mathematics Crash Course JEE Advanced>Chapter 23 - Definite Integration>Exercise 1>Q 45

If

A (x + y) = A (x) A (y) \forall x, y

and

A (0) \neq 0

and

B (x) = \frac{A (x)}{1 + {(A (x))}^{2}}

, then

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Mathematics

IMPORTANT

Mathematics Crash Course JEE Advanced>Chapter 23 - Definite Integration>Exercise 1>Q 46

Let

f (x) = \int_{0}^{x} \frac{e^{t}}{t} d t (x > 0)

, then

e^{- a} [f (x + a) - f (1 + a)] =

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Mathematics

IMPORTANT

Mathematics Crash Course JEE Advanced>Chapter 23 - Definite Integration>Exercise 1>Q 47

Given

f

is an odd function defined everywhere, periodic with period

2

and integrable on every interval. Let

g (x) = \int_{0}^{x} f (t) d t

, then

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Mathematics

IMPORTANT

Mathematics Crash Course JEE Advanced>Chapter 23 - Definite Integration>Exercise 1>Q 48

Let

I_{n} = \int_{0}^{\frac{1}{2}} \frac{1}{\sqrt{1 - x^{n}}} d x

where

n > 2

, then

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Mathematics

IMPORTANT

Mathematics Crash Course JEE Advanced>Chapter 23 - Definite Integration>Exercise 1>Q 49

I_{1} = \int_{0}^{π} \frac{x \sin x}{1 + \cos^{2} x} d x, I_{2} = \int_{0}^{π} \frac{x^{3} \sin x}{(π^{2} - 3 π x + 3 x^{2}) (1 + \cos^{2} x)} d x,

then

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Mathematics

IMPORTANT

Mathematics Crash Course JEE Advanced>Chapter 23 - Definite Integration>Exercise 1>Q 50

The value of integral

\int_{0}^{π} x f (\sin x) d x =

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Mathematics

IMPORTANT

Mathematics Crash Course JEE Advanced>Chapter 23 - Definite Integration>Exercise 1>Q 51

Let

f (x) = \int_{1}^{x} \frac{3^{t}}{1 + t^{2}} d t, x > 0,

then

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Mathematics

IMPORTANT

Mathematics Crash Course JEE Advanced>Chapter 23 - Definite Integration>Exercise 1>Q 52

Let

ϕ (x, t) = \{\begin{cases} x (t - 1), x \leq t \\ t (x - 1), t < x \end{cases}

, where

t

is a continuous function of

x

in

[0, 1]

. Let

g (x) = \int_{0}^{1} f (t) ϕ (x, t) d t;

then

g^{"} (x) =