Question

The value of the integral \[\int\limits_0^1 {{e^{{x^2}}}} dx\] is:
A. Less than \[e\]
B. Greater than \[e\]
C. Less than 1
D. Greater than 1

Answer 1

Hint: We will first let the given integral as \[I = \int\limits_0^1 {{e^{{x^2}}}} dx\]. Now, as we are given the range of integral from 0 to 1 then for \[x\] also, the range is from 0 to 1. Also, \[{x^2} < x < 1\] in the given interval because as we increase the power of \[x\] the value obtained starts decreasing so, we will use this fact and integrate the given exponential function in the given range and determine the correct option.

Complete step by step Answer:

We will first let the given integral as \[I = \int\limits_0^1 {{e^{{x^2}}}} dx\].
As we have the range from 0 to 1, then the variable \[x\] also lies within the same range and the square of \[x\] will be less than \[x\] because as we increase the power of \[x\] the value obtained will be less than the previous one.
Thus, we get,
\[ \Rightarrow {x^2} < x < 1\]
If we convert the above equation in the exponential term then this will further give us,
\[
   \Rightarrow {e^0} < {e^{{x^2}}} < {e^x} \\
   \Rightarrow 1 < {e^{{x^2}}} < {e^x} \\
 \]
Now, we will integrate the above function in the given range with respect to \[x\].
Hence, we have,
\[ \Rightarrow \int\limits_0^1 {1dx} < \int\limits_0^1 {{e^{{x^2}}}dx < } \int\limits_0^1 {{e^x}dx} \]
As we know the integration of constant results in the variable with integration is being done that is \[\int {1dx = x} \], for middle part we will substitute the value equal to \[I\] and for the third part we know that the integration of exponential function is same as exponential only that is \[\int {{e^x}dx} = {e^x}\].
Thus, we get,
\[ \Rightarrow \left[ x \right]_0^1 < I < \left[ {{e^x}} \right]_0^1\]
Next, we will apply the limits on the integrated part and we get,
\[
   \Rightarrow \left( {1 - 0} \right) < I < \left( {{e^1} - {e^0}} \right) \\
   \Rightarrow 1 < I < \left( {e - 1} \right) \\
 \]
Hence, from the above inequality, we can conclude that the value of the integral is greater than 1 and less than \[e\].
Thus, options A and D are correct.

Note: As we know that the integration of the exponential functions is equal to the exponential only such as \[\int {{e^x}dx} = {e^x}\] so, we have directly use this fact in the solution, we have written \[{x^2} < x\] in the interval 0 to 1 because as we will square the number in the given interval it will give us the lesser value that’s why \[{x^2} < x\]. Substitute the value of integral as \[I\] which directly takes us to the answer.