Let f : 0 , ∞ → R be given by f x = ∫ 1 x x e - t + 1 t d t t . Then

f 2 x is an odd function of x on R .

f x is monotonically decreasing on 0 , 1 .

Let f : 0 , ∞ → R be given by f x = ∫ 1 x x e - t + 1 t d t t . Then

f 2 x is an odd function of x on R .

The following integral ∫ π 4 π 2 2 cosec x 17 d x is equal to

∫ 0 log ( 1 + 2 ) 2 e u + e - u 16 d u

∫ 0 log ( 1 + 2 ) 2 e u + e - u 17 d u

∫ 0 log ( 1 + 2 ) 2 e u - e - u 17 d u

∫ 0 log ( 1 + 2 ) 2 e u - e - u 16 d u

E M B I B E

Alpha Question Bank for Engineering: Mathematics>Chapter 30 - Definite Integration>Exercise-3>Q 14

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Let $f : (0, \infty) \to R$ be given by $f (x) = \int_{\frac{1}{x}}^{x} e^{- (t + \frac{1}{t})} \frac{d t}{t}$ . Then

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Important Questions on Definite Integration

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Alpha Question Bank for Engineering: Mathematics>Chapter 30 - Definite Integration>Exercise-3>Q 16

The following integral

\int_{\frac{π}{4}}^{\frac{π}{2}} {(2 cosec x)}^{17} d x

is equal to

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Alpha Question Bank for Engineering: Mathematics>Chapter 30 - Definite Integration>Exercise-3>Q 19

Let $f : R \to R$ be a function defined by $x$ $f (x) = \{\begin{array}{l} [x], \\ 0, \end{array} \begin{array}{l} x \leq \sqrt{2} \\ x > \sqrt{2} \end{array}$ where $[x]$ is the greatest integer less than or equal to. If $I = \int_{- 1}^{2} \frac{x f (x^{2})}{2 + f (x + 1)} d x$ , then the value of $(4 I - 1)$ is

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Alpha Question Bank for Engineering: Mathematics>Chapter 30 - Definite Integration>Exercise-3>Q 21

Let

f : R \to R

be a continuous odd function, which vanishes exactly at one point and

f (1) = \frac{1}{2}

. Suppose that

F (x) = \int_{- 1}^{x} f (t) d t

for all

x \in [- 1, 2]

and

G (x) = \int_{- 1}^{x} t |f (f (t))| d t

for all

x \in [- 1, 2]

. If

\lim_{x \to 1} \frac{F (x)}{G (x)} = \frac{1}{14}

, then the value of

f (\frac{1}{2})

is

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Alpha Question Bank for Engineering: Mathematics>Chapter 30 - Definite Integration>Exercise-3>Q 22

Let $f (x) = 7 \tan^{8} x + 7 \tan^{6} x - 3 \tan^{4} x - 3$ $\tan^{2} x$ for all $x \in (- \frac{π}{2}, \frac{π}{2})$ . Then, the correct expression(s) is/are