the value of int0 pow 1 1 e pow x pow 2 dx is 767022074

The value of $$\int_0^1 {\left( {1 + {e^{ - {x^2}}}} \right)dx} $$ is :

A. $$-1$$

B. 2

C. $$1 + {e^{ - 1}}$$

D. none of these

Answer : none of these

Solution :
$$\eqalign{ & I = \left[ x \right]_0^1 + \int_0^1 {\frac{1}{{{e^{{x^2}}}}}dx} = 1 + \int_0^1 {\frac{1}{{{e^{{x^2}}}}}} dx \cr & {\text{In }}\left( {0,\,1} \right),\,\frac{1}{e} < \frac{1}{{{e^{{x^2}}}}} < 1 \cr & \therefore \int_0^1 {\frac{1}{e}dx < \int_0^1 {\frac{1}{{{e^{{x^2}}}}}dx < } \int_0^1 {1\,dx} } \cr & \therefore \frac{1}{e} < \int_0^1 {\frac{1}{{{e^{{x^2}}}}}dx} < 1 \cr & \therefore 1 + \frac{1}{e} < \int_0^1 {\left( {1 + {e^{ - {x^2}}}} \right)dx} < 1 + 1 \cr} $$

Releted MCQ Question on
Calculus >> Application of Integration

Releted Question 1

The area bounded by the curves $$y = f\left( x \right),$$ the $$x$$-axis and the ordinates $$x = 1$$ and $$x = b$$ is $$\left( {b - 1} \right)\sin \left( {3b + 4} \right).$$ Then $$f\left( x \right)$$ is-

A. $$\left( {x - 1} \right)\cos \left( {3x + 4} \right)$$

B. $$\sin \,\left( {3x + 4} \right)$$

C. $$\sin \,\left( {3x + 4} \right) + 3\left( {x - 1} \right)\cos \left( {3x + 4} \right)$$

D. none of these

View Answer

Releted Question 2

The area bounded by the curves $$y = \left| x \right| - 1$$ and $$y = - \left| x \right| + 1$$ is-

A. $$1$$

B. $$2$$

C. $$2\sqrt 2 $$

D. $$4$$

View Answer

Releted Question 3

The area bounded by the curves $$y = \sqrt x ,\,2y + 3 = x$$ and $$x$$-axis in the 1^st quadrant is-

A. $$9$$

B. $$\frac{{27}}{4}$$

C. $$36$$

D. $$18$$

View Answer

Releted Question 4

The area enclosed between the curves $$y = a{x^2}$$ and $$x = a{y^2}\left( {a > 0} \right)$$ is 1 sq. unit, then the value of $$a$$ is-

A. $$\frac{1}{{\sqrt 3 }}$$

B. $$\frac{1}{2}$$

C. $$1$$

D. $$\frac{1}{3}$$

View Answer

The value of $$\int_0^1 {\left( {1 + {e^{ - {x^2}}}} \right)dx} $$ is :

Releted MCQ Question on
Calculus >> Application of Integration

The area bounded by the curves $$y = f\left( x \right),$$ the $$x$$-axis and the ordinates $$x = 1$$ and $$x = b$$ is $$\left( {b - 1} \right)\sin \left( {3b + 4} \right).$$ Then $$f\left( x \right)$$ is-

The area bounded by the curves $$y = \left| x \right| - 1$$ and $$y = - \left| x \right| + 1$$ is-

The area bounded by the curves $$y = \sqrt x ,\,2y + 3 = x$$ and $$x$$-axis in the 1^st quadrant is-

The area enclosed between the curves $$y = a{x^2}$$ and $$x = a{y^2}\left( {a > 0} \right)$$ is 1 sq. unit, then the value of $$a$$ is-

Practice More Releted MCQ Question on
Application of Integration

Practice More MCQ Question on Maths Section

The value of $$\int_0^1 {\left( {1 + {e^{ - {x^2}}}} \right)dx} $$ is :

Releted MCQ Question on Calculus >> Application of Integration

The area bounded by the curves $$y = f\left( x \right),$$ the $$x$$-axis and the ordinates $$x = 1$$ and $$x = b$$ is $$\left( {b - 1} \right)\sin \left( {3b + 4} \right).$$ Then $$f\left( x \right)$$ is-

The area bounded by the curves $$y = \left| x \right| - 1$$ and $$y = - \left| x \right| + 1$$ is-

The area bounded by the curves $$y = \sqrt x ,\,2y + 3 = x$$ and $$x$$-axis in the 1st quadrant is-

The area enclosed between the curves $$y = a{x^2}$$ and $$x = a{y^2}\left( {a > 0} \right)$$ is 1 sq. unit, then the value of $$a$$ is-

Practice More Releted MCQ Question on Application of Integration

Practice More MCQ Question on Maths Section

Releted MCQ Question on
Calculus >> Application of Integration

The area bounded by the curves $$y = \sqrt x ,\,2y + 3 = x$$ and $$x$$-axis in the 1^st quadrant is-

Practice More Releted MCQ Question on
Application of Integration