integration for integrators

If P (x) is a polynomial of least degree that has a maximum equal to 6 at x = 1 , and a minimum equal to 2 at
x = 3 , then01∫ P(x) dx equals

1. 17/4

2. 13/4

3. 19/4

4. 5/4

A point 'P' moves in xy plane in such a way that [| x | + | y |] = 1, where [.] denotes the greatest integer function. Area of the region representing all possible positions of the point 'P' is equal to

1. 4 sq units

2. 16 sq units

3. 2 2 sq units

4. 8 sq units

The area enclosed between the curve y = log e (x + e) and the coordinate axes is

1.] 2

2.] 1

3.] 4

4.] 3

42 Answers

106
Asish Mahapatra ·

3rd answer is area is 1

62
Lokesh Verma ·

y = ln(x + e)

where ln stands for base e

so now x=-e we have y going to - infinity

also at x=0,

so the integral will be from -e to 0 of

ln(x+e) which is same as integral from 0 to e of ln(x)

= x ln(x/e) | limti from 0 to e

you have to take the RHL when x approaches zero!!!

is this hint sufficient?

11
Mani Pal Singh ·

koi isko dekho to sahi!!!!!!!!!!!

11
Mani Pal Singh ·

what about 1st sabse mast hai na!!!!!!
nishant sir
jo aapne bataya uss mein koi prob nahin hai but pls tell me what is the area bounded by the ex from -∞ to ∞

62
Lokesh Verma ·

that area is not defined manipal.

62
Lokesh Verma ·

haan 1st is the best one :)

11
Mani Pal Singh ·

so what is the logic of 2nd ques

62
Lokesh Verma ·

ashish;s solution is perfect.. check that pinked post

11
Mani Pal Singh ·

arrey pls reply
these will help us

62
Lokesh Verma ·

logic of 2nd question..

it is symmetric in the 4 quadrants

we take the 1st quadrant

there the graph becomes

[x+y]=1

this is same as

1<=x+y<2

now plot this and find the area.. multiply this by 4

11
Mani Pal Singh ·

what about 1st
3rd ka logic bhi ek bar phi se samjha do

1
Terminator ·

sir in 1st quadrant is 1<=x+y<2

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