Maths-

General

Easy

Question

The value of the definite integral $stretchy integral subscript 0 end subscript superscript 1 end superscript open parentheses 1 plus e to the power of negative x to the power of 2 end exponent end exponent close parentheses d x$ is

1
2
$1 + e^{1}$
noneofthese

hint Hint:

We are aware that differentiation is the process of discovering a function's derivative and integration is the process of discovering a function's antiderivative. Thus, both processes are the antithesis of one another. Therefore, we can say that differentiation is the process of differentiation and integration is the reverse. The anti-differentiation is another name for the integration.
Here we have given: $stretchy integral subscript 0 end subscript superscript 1 end superscript open parentheses 1 plus e to the power of negative x to the power of 2 end exponent end exponent close parentheses d x$ and we have to integrate it. We will use minimum and maximum method to solve this.

The correct answer is: noneofthese

Now we have given the function as $stretchy integral subscript 0 superscript 1 open parentheses 1 plus e to the power of negative x squared end exponent close parentheses d x$ . Here the lower limit is 0 and upper limit is 1. We know that there are some integrals where we can use the minimum and maximum method which makes problems to solve easily. We will use one of them.
We have:
$stretchy integral subscript 0 superscript 1 open parentheses 1 plus e to the power of negative x squared end exponent close parentheses d x N o w space c o n s i d e r space t h e space f u n n c t i o n space f left parenthesis x right parenthesis comma space i f space i t space i s space c o n t i n u o u s space f u n c t i o n space i n space a comma b space t h e n colon m left parenthesis b minus a right parenthesis less or equal than stretchy integral subscript a superscript b f left parenthesis x right parenthesis d x less or equal than M left parenthesis b minus a right parenthesis H e r e colon M equals space m a x i m i m space v a l u e space o f space f left parenthesis x right parenthesis space i n space left square bracket a comma b right square bracket m equals space m i n i m u m space v a l u e space o f space f left parenthesis x right parenthesis space i n space left square bracket a comma b right square bracket N o w space t h e space f u n c t i o n space w e space h a v e space i s colon f left parenthesis x right parenthesis equals open parentheses 1 plus e to the power of negative x squared end exponent close parentheses space I t space i s space c o n t i n u o u s space i n space left square bracket 0 comma 1 right square bracket. T h e space l i m i t s space a r e space f r o m space 0 space t o space 1 comma space s o colon i f space 0 space less than space x space less than space 1 comma space t h e n x squared less than x e to the power of x squared end exponent less than e to the power of x e to the power of negative x squared end exponent less than e to the power of negative x end exponent i f space 0 space less than space x space less than space 1 comma space t h e n x squared greater than 0 e to the power of x squared end exponent greater than e to the power of 0 e to the power of negative x squared end exponent less than 1 S o space w e space g e t colon e to the power of negative x end exponent less than e to the power of negative x squared end exponent less than 1 space f o r space a l l space x element of left square bracket 0 comma 1 right square bracket. N o w space a d d i n g space 1 space t o space a l l space t h r e e space t e r m s comma space w e space g e t colon 1 plus e to the power of negative x end exponent less than 1 plus e to the power of negative x squared end exponent less than 1 plus 1 A d d i n g space i n t e g r a l space s i g n comma space w e space g e t colon stretchy integral subscript 0 superscript 1 left parenthesis 1 plus e to the power of negative x end exponent right parenthesis d x less than stretchy integral subscript 0 superscript 1 left parenthesis 1 plus e to the power of negative x squared end exponent right parenthesis d x less than stretchy integral subscript 0 superscript 1 2 d x A f t e r space p u t t i n g space t h e space l i m i t s comma space w e space g e t colon 2 minus 1 over e less than stretchy integral subscript 0 superscript 1 left parenthesis 1 plus e to the power of negative x squared end exponent right parenthesis d x less than 2$

So here we used the concept of integrals and simplified it. We can also solve it manually but it will take lot of time to come to final answer hence we used the minimum and maximum method which makes problem to solve easily. So the answer is non of these.

Have a query? Ask a Tutor!!

Can’t find what you’re looking for?

The value of the definite integral is

1 2 noneofthese

The correct answer is: noneofthese

Now we have given the function as . Here the lower limit is 0 and upper limit is 1. We know that there are some integrals where we can use the minimum and maximum method which makes problems to solve easily. We will use one of them.We have:

Related Questions to study

Suppos ef, f' andf'' are continuous on[0, e] and that then the value of

Suppos ef, f' andf'' are continuous on[0, e] and that then the value of

ther

ther

If is a point on the circle whose centre is on the x-axis and which touches the line x+y=0 at (2,-2), then the greatest value of a is

If is a point on the circle whose centre is on the x-axis and which touches the line x+y=0 at (2,-2), then the greatest value of a is

Equation of the circle of radius is

Equation of the circle of radius is

In the adjoining figure, the circle meets the sides of an equilateral triangle at six points. If AG=2,GF=13,FC=1 and HJ=7, then DE=

In the adjoining figure, the circle meets the sides of an equilateral triangle at six points. If AG=2,GF=13,FC=1 and HJ=7, then DE=

Two circles of unequal radii intersect in two distinct points P and Q. A line through P cuts 1st circle at A and 2nd circle at B. Let M be the mid point of AB. If the line MQ meets the 1st circle at X and the second circle at Z then M divides XZ internally in the ratio.

Two circles of unequal radii intersect in two distinct points P and Q. A line through P cuts 1st circle at A and 2nd circle at B. Let M be the mid point of AB. If the line MQ meets the 1st circle at X and the second circle at Z then M divides XZ internally in the ratio.

A circle touches the lines and has unit radius. If the centre of this circle lies in the first quadrant, then one possible equation of this circle is

A circle touches the lines and has unit radius. If the centre of this circle lies in the first quadrant, then one possible equation of this circle is

From a point R(5, 8) two tangents RP and RQ are drawn to a given circle S = 0 whose radius is 5. If circumcentre of the triangle PQR is (2, 3), then the equation of circle S = 0 is

From a point R(5, 8) two tangents RP and RQ are drawn to a given circle S = 0 whose radius is 5. If circumcentre of the triangle PQR is (2, 3), then the equation of circle S = 0 is

PA and PB are tangents to a circle S touching S at points A and B.C is a point on S in between A and B as shown in the figure. LCM is a tangent to S intersecting PA and PB in points L and M respectively.Then the perimeter of the triangle PLM depends on

PA and PB are tangents to a circle S touching S at points A and B.C is a point on S in between A and B as shown in the figure. LCM is a tangent to S intersecting PA and PB in points L and M respectively.Then the perimeter of the triangle PLM depends on

is a point on circle locate the points on the given circle which are at 2 unit distance from

is a point on circle locate the points on the given circle which are at 2 unit distance from

In the adjoining figure, the circle meets the sides of an equilateral triangle at six points. If Ag = 2, GF = 13, FC = 1 and HJ = 7, then DE equals

In the adjoining figure, the circle meets the sides of an equilateral triangle at six points. If Ag = 2, GF = 13, FC = 1 and HJ = 7, then DE equals

If a line segment AM = a moves in the plane XOY remaining parallel to OX so that the left end point A slides along the circle

If a line segment AM = a moves in the plane XOY remaining parallel to OX so that the left end point A slides along the circle

The triangle PQR is inscribed in the circle If Q and R have coordinates (3, 4) and (-4, 3) respectively, then is

The triangle PQR is inscribed in the circle If Q and R have coordinates (3, 4) and (-4, 3) respectively, then is

For all values of m ÎR the line y-mx + m-1 = 0 cuts the circle at an angle

For all values of m ÎR the line y-mx + m-1 = 0 cuts the circle at an angle

is a circle of radius 1 touching the x -axis and y -axis. is another circle of radius and touching the axes as well as circle

is a circle of radius 1 touching the x -axis and y -axis. is another circle of radius and touching the axes as well as circle

With Turito Academy.

With Turito Foundation.

Get an Expert Advice From Turito.

Turito Academy

With Turito Academy.

Test Prep

With Turito Foundation.

Courses

Quick Links

The value of the definite integral $stretchy integral subscript 0 end subscript superscript 1 end superscript open parentheses 1 plus e to the power of negative x to the power of 2 end exponent end exponent close parentheses d x$ is

1
2
$1 + e^{1}$
noneofthese

If $left parenthesis alpha beta comma right parenthesis$ is a point on the circle whose centre is on the x-axis and which touches the line x+y=0 at (2,-2), then the greatest value of a is

If $left parenthesis alpha beta comma right parenthesis$ is a point on the circle whose centre is on the x-axis and which touches the line x+y=0 at (2,-2), then the greatest value of a is

Equation of the circle of radius $square root of 2 text ´ containing the point end text left parenthesis 3 , 1 right parenthesis text and touching the line end text vertical line x minus 1 vertical line equals vertical line y minus 1 vertical line$ is

Equation of the circle of radius $square root of 2 text ´ containing the point end text left parenthesis 3 , 1 right parenthesis text and touching the line end text vertical line x minus 1 vertical line equals vertical line y minus 1 vertical line$ is

A circle touches the lines $y equals fraction numerator x over denominator square root of 3 end fraction comma y equals x square root of 3$ and has unit radius. If the centre of this circle lies in the first quadrant, then one possible equation of this circle is

A circle touches the lines $y equals fraction numerator x over denominator square root of 3 end fraction comma y equals x square root of 3$ and has unit radius. If the centre of this circle lies in the first quadrant, then one possible equation of this circle is

$left parenthesis 1 , 2 square root of 2 right parenthesis$ is a point on circle $x to the power of 2 end exponent plus y to the power of 2 end exponent equals 9$ locate the points on the given circle which are at 2 unit distance from $left parenthesis 1 , 2 square root of 2 right parenthesis$

$left parenthesis 1 , 2 square root of 2 right parenthesis$ is a point on circle $x to the power of 2 end exponent plus y to the power of 2 end exponent equals 9$ locate the points on the given circle which are at 2 unit distance from $left parenthesis 1 , 2 square root of 2 right parenthesis$

If a line segment AM = a moves in the plane XOY remaining parallel to OX so that the left end point A slides along the circle $x to the power of 2 end exponent plus y to the power of 2 end exponent equals a to the power of 2 end exponent comma text the locus of end text M text is end text$

If a line segment AM = a moves in the plane XOY remaining parallel to OX so that the left end point A slides along the circle $x to the power of 2 end exponent plus y to the power of 2 end exponent equals a to the power of 2 end exponent comma text the locus of end text M text is end text$

The triangle PQR is inscribed in the circle $x to the power of 2 end exponent plus y to the power of 2 end exponent equals 25 text . end text$ If Q and R have coordinates (3, 4) and (-4, 3) respectively, then $angle Q P R$ is

The triangle PQR is inscribed in the circle $x to the power of 2 end exponent plus y to the power of 2 end exponent equals 25 text . end text$ If Q and R have coordinates (3, 4) and (-4, 3) respectively, then $angle Q P R$ is

For all values of m ÎR the line y-mx + m-1 = 0 cuts the circle $x to the power of 2 end exponent plus y to the power of 2 end exponent minus 2 x minus 2 y plus 1 equals 0 text end text$ at an angle

For all values of m ÎR the line y-mx + m-1 = 0 cuts the circle $x to the power of 2 end exponent plus y to the power of 2 end exponent minus 2 x minus 2 y plus 1 equals 0 text end text$ at an angle