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The value of the integral \int_{0}^{1}e^{x^{2}}dx is

  • Option 1)

    less than e

  • Option 2)

    greater than e

  • Option 3)

    less than 1

  • Option 4)

    greater than 2

 

Answers (1)

As we learnt

Properties of Definite Integration -

If m \, and\, M are smallest and greatest value of a function f(x) defined on an interval (a, b ) then

m\left ( b-a \right )\leq \int_{a}^{b}f(x)dx\leq M\left ( b-a \right )
 

- wherein

since

m\leq f(x)\leq M

so

\int_{a}^{b}mdx\leq \int_{a}^{b}f(x)dx\leq \int_{a}^{b}Mdx

 

 

For 0 < x < 1, we have 1 < e^{x^{2}} < e, so that

                        \int_{0}^{1}1dx< \int_{0}^{1}e^{x^{2}}dx< \int_{0}^{1}edx\; \therefore 1< \int_{0}^{1}e^{x^{2}}dx<e 


Option 1)

less than e

Option 2)

greater than e

Option 3)

less than 1

Option 4)

greater than 2

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subam

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