Question

Find the value of the definite integral \[\int\limits_0^1 {\left( {1 + {e^{ - {x^2}}}} \right)} dx\].
A. \[ - 1\]
B. 2
C. \[1 + {e^{ - 1}}\]
D. None of these

Answer 1

Hint: Here, a definite integral is given. First, use the sum rule of the integration \[\int\limits_a^b {\left[ {f\left( x \right) + g\left( x \right)} \right]} dx = \int\limits_a^b {f\left( x \right)} dx + \int\limits_a^b {g\left( x \right)} dx\] and split the terms of the given integral. Then, solve the integrals by applying the integration formulas.



Formula Used:Integration Formula: \[\int\limits_a^b {ndx = \left[ {nx} \right]} _a^b = n\left( {b - a} \right)\]
The sum rule of the integration: \[\int\limits_a^b {\left[ {f\left( x \right) + g\left( x \right)} \right]} dx = \int\limits_a^b {f\left( x \right)} dx + \int\limits_a^b {g\left( x \right)} dx\]



Complete step by step solution:The given definite integral is \[\int\limits_0^1 {\left( {1 + {e^{ - {x^2}}}} \right)} dx\].
Let consider,
\[I = \int\limits_0^1 {\left( {1 + {e^{ - {x^2}}}} \right)} dx\]
Apply the sum rule of the integration \[\int\limits_a^b {\left[ {f\left( x \right) + g\left( x \right)} \right]} dx = \int\limits_a^b {f\left( x \right)} dx + \int\limits_a^b {g\left( x \right)} dx\].
\[I = \int\limits_0^1 1 dx + \int\limits_0^1 {{e^{ - {x^2}}}} dx\]
Solve the first integral by applying the formula \[\int\limits_a^b {ndx = \left[ {nx} \right]} _a^b = n\left( {b - a} \right)\].
\[I = \left[ x \right]_0^1 + \int\limits_0^1 {{e^{ - {x^2}}}} dx\]
\[ \Rightarrow I = \left( {1 - 0} \right) + \int\limits_0^1 {{e^{ - {x^2}}}} dx\]
\[ \Rightarrow I = 1 + \int\limits_0^1 {{e^{ - {x^2}}}} dx\]
But the second term \[\int\limits_0^1 {{e^{ - {x^2}}}} dx\] is not integrable.
Therefore, the value of the integral \[\int\limits_0^1 {\left( {1 + {e^{ - {x^2}}}} \right)} dx\] does not exists.



Option ‘D’ is correct



Note: Students often get confused and solve the second integral \[\int\limits_0^1 {{e^{ - {x^2}}}} dx\] by applying the integration formula \[\int {{e^x}} dx = {e^x}\] . Because of that, they get the value of the integrals as \[\int\limits_0^1 {{e^{ - {x^2}}}} dx = \dfrac{1}{e} - 1\] and \[\int\limits_0^1 {\left( {1 + {e^{ - {x^2}}}} \right)} dx = 1 + \dfrac{1}{e} - 1 = \dfrac{1}{e}\]. But both values are wrong, so they get the wrong solution.