Question

Find the interval where the values of the definite integral \[\int\limits_0^1 {{e^{{x^2}}}} dx\] lies.
A. \[\left( {0,1} \right)\]
B. \[\left( { - 1,0} \right)\]
C. \[\left( {1,e} \right)\]
D. None of these

Answer 1

Hint: Here, a definite integral of the exponential function is given. First, calculate the minimum and maximum values of the function \[{e^{{x^2}}}\] in the interval \[\left( {0,1} \right)\] and write them in the form of inequality. Then, integrate the inequality equation with respect to the variable \[x\]. After that, solve the integrals and find the required value of the integral.



Formula Used:Integration Formula: \[\int\limits_a^b {ndx = \left[ {nx} \right]} _a^b = n\left( {b - a} \right)\]
\[\int\limits_a^b n f\left( x \right)dx = n\int\limits_a^b {f\left( x \right)dx} \]



Complete step by step solution:The given definite integral is \[\int\limits_0^1 {{e^{{x^2}}}} dx\].
Since \[{e^{{x^2}}}\] is an exponential function. So, it is increasing in the interval \[\left( {0,1} \right)\].
Now calculate the minimum and maximum values of the function \[{e^{{x^2}}}\] in the interval \[\left( {0,1} \right)\].
Minimum value: \[{e^{{0^2}}} = {e^0} = 1\]
Maximum value: \[{e^{{1^2}}} = {e^1} = e\]
 So, we get
\[1 < {e^{{x^2}}} < e\]
Now integrate the inequality equation with respect to the variable \[x\].
\[\int\limits_0^1 1 dx < \int\limits_0^1 {{e^{{x^2}}}} dx < \int\limits_0^1 e dx\]
Apply the constant property of the integration \[\int\limits_a^b n f\left( x \right)dx = n\int\limits_a^b {f\left( x \right)dx} \].
\[ \Rightarrow \int\limits_0^1 {dx} < \int\limits_0^1 {{e^{{x^2}}}} dx < e\int\limits_0^1 {dx} \]
Solve the first and third integral by applying the integration rule \[\int\limits_a^b {ndx = \left[ {nx} \right]} _a^b = n\left( {b - a} \right)\].
\[ \Rightarrow \left[ x \right]_0^1 < \int\limits_0^1 {{e^{{x^2}}}} dx < e\left[ x \right]_0^1\]
Apply the upper and lower limits.
\[ \Rightarrow \left( {1 - 0} \right) < \int\limits_0^1 {{e^{{x^2}}}} dx < e\left( {1 - 0} \right)\]
\[ \Rightarrow 1 < \int\limits_0^1 {{e^{{x^2}}}} dx < e\]
The minimum and the maximum values of the integral \[\int\limits_0^1 {{e^{{x^2}}}} dx\] are 1 and \[e\] respectively.
Therefore, the values of the definite integral \[\int\limits_0^1 {{e^{{x^2}}}} dx\] lies in the interval \[\left( {1,e} \right)\].



Option ‘C’ is correct



Note: students often get confused and solve the given integral by using the formula \[\int {{e^x}dx = {e^x}} \]. Because of that, they get the wrong answer.