Newton-Leibniz & Reduction Formulas

Interactive Feed

JEE Main 2013

The intercepts on $x$ -axis made by tangents to the curve, $y = \int_{0}^{x} ∣ t ∣ d t, x \in R$ , which are parallel to the line $y = 2 x$ , are equal to

\pm 1

(B)

\pm 2

(C)

\pm 3

(D)

\pm 4

JEE Main 2011

LEVELJEE Main

For $x \in (0, 5 π /2)$ , define $f (x) = \int_{0}^{x} t sin t d t$ . Then $f$ has

(A)

local minimum at

π

and

2 π

(B)

local minimum at

π

and local maximum at

2 π

(C)

local maximum at

π

and local minimum at

2 π

(D)

local maximum at

π

and

2 π

JEE Main 2007

LEVELJEE Main

Let $F (x) = f (x) + f (1/ x)$ , where $f (x) = \int_{1}^{x} \frac{l o g t}{1 + t} d t$ . Then $F (e)$ equals

Let $f : R \to R$ be a differentiable function having $f (2) = 6, f^{'} (2) = \frac{1}{48}$ . Then $lim_{x \to 2} \int_{6}^{f (x)} \frac{4 t ^{3}}{x - 2} d t$ equals

(A)

24

(B)

36

(C)

12

(D)

18

JEE Main 2002

LEVELJEE Main

$I_{n} = \int_{0}^{π /4} tan^{n} x d x$ then $lim_{n \to \infty} n [I_{n} + I_{n + 2}]$ equals

\infty

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

* Multiple Correct Options

(A)

f (x)

is monotonically increasing on

[1, \infty)

(B)

f (x)

is monotonically decreasing on

(0, 1)

(C)

f (x) + f (1/ x) = 0

, for all

x \in (0, \infty)

(D)

f (2^{x})

is an odd function of

x

R

JEE Advanced 2009

LEVELJEE Advanced

Let $f$ be a non-negative function defined on the interval [0, 1]. If $\int_{0}^{x} 1 - (f^{'} (t))^{2} d t = \int_{0}^{x} f (t) d t$ , $0 \leq x \leq 1$ , and $f (0) = 0$ , then

(A)

f (1/2) < 1/2

and

f (1/3) > 1/3

(B)

f (1/2) > 1/2

and

f (1/3) > 1/3

(C)

f (1/2) < 1/2

and

f (1/3) < 1/3

(D)

f (1/2) > 1/2

and

f (1/3) < 1/3

JEE Advanced 2007

LEVELJEE Main

$lim_{x \to π /4} \frac{\int _{2}^{s e c^{2} x} f ( t ) d t}{x ^{2} - \frac{π ^{2}}{16}}$ equals

(A)

\frac{8}{π} f (2)

(B)

\frac{2}{π} f (2)

(C)

\frac{2}{π} f (1/2)

(D)

4 f (2)

JEE Advanced 2005

LEVELJEE Main

If $\int_{s i n x}^{1} t^{2} f (t) d t = 1 - sin x$ , then $f (\frac{1}{3})$ is

(A)

1/3

(B)

1/ 3

(C)

3

(D)

3

JEE Advanced 2004

LEVELJEE Main

If $f (x)$ is differentiable and $\int_{0}^{t^{2}} x f (x) d x = \frac{2}{5} t^{5}$ , then $f (4/25)$ equals

Let $f (x) = \int_{1}^{x} 2 - t^{2} d t$ . Then the real roots of the equation $x^{2} - f^{'} (x) = 0$ are

(A)

\pm 1

(B)

\pm \frac{1}{2}

(C)

\pm \frac{1}{2}

Let $f : (0, \infty) \to R$ and $F (x) = \int_{0}^{x} f (t) d t$ . If $F (x^{2}) = x^{2} (1 + x)$ , then $f (4)$ equals

(A)

5/4

(B)

7

(C)

4

(D)

2

JEE Advanced 1998

LEVELJEE Main

If $\int_{0}^{x} f (t) d t = x + \int_{x}^{1} t f (t) d t$ , then the value of $f (1)$ is

Let $f : R \to R$ be a differentiable function and $f (1) = 4$ . Then the value of $lim_{x \to 1} \int_{4}^{f (x)} \frac{2 t}{x - 1} d t$ is

(A)

8 f^{'} (1)

(B)

4 f^{'} (1)

(C)

2 f^{'} (1)

(D)

f^{'} (1)

JEE Advanced 2016

LEVELJEE Main

The total number of distinct $x \in [0, 1]$ for which $\int_{0}^{x} \frac{t ^{2}}{1 + t ^{4}} d t = 2 x - 1$ is

JEE Advanced 2015

LEVELJEE Advanced

Let $F (x) = \int_{x}^{x^{2} + \frac{π}{6}} 2 cos^{2} t (d t)$ for all $x \in R$ and $f : [0, 1/2] \to [0, \infty)$ be a continuous function. For $α \in [0, 1/2]$ , if $F^{'} (α) + 2$ is the area of the region bounded by $x = 0, y = 0, y = f (x)$ and $x = α$ , then $f (0)$ is

JEE Advanced 2015

LEVELJEE Advanced

Let $f : R \to R$ be a continuous odd function, which vanishes exactly at one point and $f (1) = 1/2$ . 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 $f (1/2)$ is

JEE Advanced 2006

LEVELJEE Main

The value of $5050 \frac{\int _{0}^{1} ( 1 - x ^{50} ) ^{100} d x}{\int _{0}^{1} ( 1 - x ^{50} ) ^{101} d x}$ is.

JEE Advanced 1992

LEVELJEE Main

Determine a positive integer $n \leq 5$ , such that $\int_{0}^{1} e^{x} (x - 1)^{n} d x = 16 - 6 e$

JEE Advanced 2004

LEVELJEE Main

If $y (x) = \int_{π^{2} /16}^{x^{2}} \frac{c o s x c o s θ}{1 + s i n ^{2} θ} d θ$ , then find $\frac{d y}{d x}$ at $x = π$ .

JEE Advanced 1997

LEVELJEE Main

Let $a + b = 4$ , where $a < 2$ , and let $g (x)$ be a differentiable function. If $d g / d x > 0$ for all $x$ , prove that $\int_{0}^{a} g (x) d x + \int_{0}^{b} g (x) d x$ increases as $(b - a)$ increases.

JEE Advanced 1996

LEVELJEE Main

Let $A_{n}$ be the area bounded by the curve $y = (tan x)^{n}$ and the lines $x = 0, y = 0$ and $x = \frac{π}{4}$ . Prove that for $n > 2, A_{n} + A_{n - 2} = \frac{1}{n - 1}$ and deduce $\frac{1}{2 n + 2} < A_{n} < \frac{1}{2 n - 2}$ .

JEE Main 2018 (16 April Shift 1)

LEVELJEE Main

If $f (x) = \int_{0}^{x} t (sin x - sin t) d t$ then :

(A)

f^{'''} (x) + f^{'} (x) = cos x - 2 x sin x

(B)

f^{'''} (x) + f^{''} (x) - f^{'} (x) = cos x

(C)

f^{'''} (x) - f^{''} (x) = cos x - 2 x sin x

(D)

f^{'''} (x) + f^{''} (x) = sin x

JEE Main 2019 (10 January)

LEVELJEE Main

If $\int_{0}^{x} f (t) d t = x^{2} + \int_{x}^{1} t^{2} f (t) d t$ , then $f^{'} (1/2)$ is :

(A)

\frac{6}{25}

(B)

\frac{24}{25}

(C)

\frac{18}{25}

(D)

\frac{4}{5}

JEE Main 2019 (9 April)

LEVELJEE Main

If $f : R \to R$ is a differentiable function and $f (2) = 6$ , then $lim_{x \to 2} \frac{\int _{6}^{f (x)} 2 t d t}{x - 2}$ is :-

(A)

(B)

2 f^{'} (2)

(C)

12 f^{'} (2)

(D)

24 f^{'} (2)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The intercepts on x-axis made by tangents to the curve, y=∫0x​∣t∣dt,x∈R, which are parallel to the line y=2x, are equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For x∈(0,5π/2), define f(x)=∫0x​t​sintdt. Then f has

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let F(x)=f(x)+f(1/x), where f(x)=∫1x​1+tlogt​dt. Then F(e) equals

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f:R→R be a differentiable function having f(2)=6,f′(2)=481​. Then limx→2​∫6f(x)​x−24t3​dt equals

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } In​=∫0π/4​tannxdx then limn→∞​n[In​+In+2​] equals

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f:(0,∞)→R be given by f(x)=∫1/xx​e−(t+1/t)tdt​. Then

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f be a non-negative function defined on the interval [0, 1]. If ∫0x​1−(f′(t))2​dt=∫0x​f(t)dt, 0≤x≤1, and f(0)=0, then

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } limx→π/4​x2−16π2​∫2sec2x​f(t)dt​ equals

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If ∫sinx1​t2f(t)dt=1−sinx, then f(3​1​) is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f(x) is differentiable and ∫0t2​xf(x)dx=52​t5, then f(4/25) equals

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x)=∫1x​2−t2​dt. Then the real roots of the equation x2−f′(x)=0 are

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f:(0,∞)→R and F(x)=∫0x​f(t)dt. If F(x2)=x2(1+x), then f(4) equals

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If ∫0x​f(t)dt=x+∫x1​tf(t)dt, then the value of f(1) is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f:R→R be a differentiable function and f(1)=4. Then the value of limx→1​∫4f(x)​x−12t​dt is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The total number of distinct x∈[0,1] for which ∫0x​1+t4t2​dt=2x−1 is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of 5050∫01​(1−x50)101dx∫01​(1−x50)100dx​ is.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Determine a positive integer n≤5, such that ∫01​ex(x−1)ndx=16−6e

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If y(x)=∫π2/16x2​1+sin2θ​cosxcosθ​​dθ, then find dxdy​ at x=π.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a+b=4, where a<2, and let g(x) be a differentiable function. If dg/dx>0 for all x, prove that ∫0a​g(x)dx+∫0b​g(x)dx increases as (b−a) increases.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let An​ be the area bounded by the curve y=(tanx)n and the lines x=0,y=0 and x=4π​. Prove that for n>2,An​+An−2​=n−11​ and deduce 2n+21​<An​<2n−21​.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f(x)=∫0x​t(sinx−sint)dt then :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If ∫0x​f(t)dt=x2+∫x1​t2f(t)dt, then f′(1/2) is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f:R→R is a differentiable function and f(2)=6, then limx→2​x−2∫6f(x)​2tdt​ is :-

The intercepts on $x$ -axis made by tangents to the curve, $y = \int_{0}^{x} ∣ t ∣ d t, x \in R$ , which are parallel to the line $y = 2 x$ , are equal to

For $x \in (0, 5 π /2)$ , define $f (x) = \int_{0}^{x} t sin t d t$ . Then $f$ has

Let $F (x) = f (x) + f (1/ x)$ , where $f (x) = \int_{1}^{x} \frac{l o g t}{1 + t} d t$ . Then $F (e)$ equals

Let $f : R \to R$ be a differentiable function having $f (2) = 6, f^{'} (2) = \frac{1}{48}$ . Then $lim_{x \to 2} \int_{6}^{f (x)} \frac{4 t ^{3}}{x - 2} d t$ equals

$I_{n} = \int_{0}^{π /4} tan^{n} x d x$ then $lim_{n \to \infty} n [I_{n} + I_{n + 2}]$ equals

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

Let $f$ be a non-negative function defined on the interval [0, 1]. If $\int_{0}^{x} 1 - (f^{'} (t))^{2} d t = \int_{0}^{x} f (t) d t$ , $0 \leq x \leq 1$ , and $f (0) = 0$ , then

$lim_{x \to π /4} \frac{\int _{2}^{s e c^{2} x} f ( t ) d t}{x ^{2} - \frac{π ^{2}}{16}}$ equals

If $\int_{s i n x}^{1} t^{2} f (t) d t = 1 - sin x$ , then $f (\frac{1}{3})$ is

If $f (x)$ is differentiable and $\int_{0}^{t^{2}} x f (x) d x = \frac{2}{5} t^{5}$ , then $f (4/25)$ equals

Let $f (x) = \int_{1}^{x} 2 - t^{2} d t$ . Then the real roots of the equation $x^{2} - f^{'} (x) = 0$ are

Let $f : (0, \infty) \to R$ and $F (x) = \int_{0}^{x} f (t) d t$ . If $F (x^{2}) = x^{2} (1 + x)$ , then $f (4)$ equals

If $\int_{0}^{x} f (t) d t = x + \int_{x}^{1} t f (t) d t$ , then the value of $f (1)$ is

Let $f : R \to R$ be a differentiable function and $f (1) = 4$ . Then the value of $lim_{x \to 1} \int_{4}^{f (x)} \frac{2 t}{x - 1} d t$ is

The total number of distinct $x \in [0, 1]$ for which $\int_{0}^{x} \frac{t ^{2}}{1 + t ^{4}} d t = 2 x - 1$ is

The value of $5050 \frac{\int _{0}^{1} ( 1 - x ^{50} ) ^{100} d x}{\int _{0}^{1} ( 1 - x ^{50} ) ^{101} d x}$ is.

Determine a positive integer $n \leq 5$ , such that $\int_{0}^{1} e^{x} (x - 1)^{n} d x = 16 - 6 e$

If $y (x) = \int_{π^{2} /16}^{x^{2}} \frac{c o s x c o s θ}{1 + s i n ^{2} θ} d θ$ , then find $\frac{d y}{d x}$ at $x = π$ .

Let $a + b = 4$ , where $a < 2$ , and let $g (x)$ be a differentiable function. If $d g / d x > 0$ for all $x$ , prove that $\int_{0}^{a} g (x) d x + \int_{0}^{b} g (x) d x$ increases as $(b - a)$ increases.

Let $A_{n}$ be the area bounded by the curve $y = (tan x)^{n}$ and the lines $x = 0, y = 0$ and $x = \frac{π}{4}$ . Prove that for $n > 2, A_{n} + A_{n - 2} = \frac{1}{n - 1}$ and deduce $\frac{1}{2 n + 2} < A_{n} < \frac{1}{2 n - 2}$ .

If $f (x) = \int_{0}^{x} t (sin x - sin t) d t$ then :

If $\int_{0}^{x} f (t) d t = x^{2} + \int_{x}^{1} t^{2} f (t) d t$ , then $f^{'} (1/2)$ is :

If $f : R \to R$ is a differentiable function and $f (2) = 6$ , then $lim_{x \to 2} \frac{\int _{6}^{f (x)} 2 t d t}{x - 2}$ is :-