Question

The value of following integral $\int_{ - 4}^{ - 5} {{e^{{{\left( {x + 5} \right)}^2}}}dx} + \int_{\dfrac{1}{3}}^{\dfrac{2}{3}} {{e^{9{{\left( {x - \dfrac{2}{3}} \right)}^2}}}dx} $ is equal to
$\left( a \right)$ 1
$\left( b \right)$ -1
$\left( c \right)$ 0
$\left( d \right)$ -2

Answer 1

Hint: In this particular question solve the integral parts separately using integration by substitution method (I.e. change the variable by substation so that integral becomes simple don’t forget to change the limits as well), so use these concepts to reach the solution of the question.

Complete step-by-step answer:
$\int_{ - 4}^{ - 5} {{e^{{{\left( {x + 5} \right)}^2}}}dx} + \int_{\dfrac{1}{3}}^{\dfrac{2}{3}} {{e^{9{{\left( {x - \dfrac{2}{3}} \right)}^2}}}dx} $
Let, I = $\int_{ - 4}^{ - 5} {{e^{{{\left( {x + 5} \right)}^2}}}dx} + \int_{\dfrac{1}{3}}^{\dfrac{2}{3}} {{e^{9{{\left( {x - \dfrac{2}{3}} \right)}^2}}}dx} $
Let, ${I_1} = \int_{ - 4}^{ - 5} {{e^{{{\left( {x + 5} \right)}^2}}}dx} $
And, ${I_2} = \int_{\dfrac{1}{3}}^{\dfrac{2}{3}} {{e^{9{{\left( {x - \dfrac{2}{3}} \right)}^2}}}dx} $
Therefore, $I = {I_1} + {I_2}$.................. (1)
So first simplify first integral we have,
$ \Rightarrow {I_1} = \int_{ - 4}^{ - 5} {{e^{{{\left( {x + 5} \right)}^2}}}dx} $.................... (2)
In the above integral substitute (x + 5) = t............ (3)
Now differentiate equation (3) w.r.t x we have,
$ \Rightarrow \dfrac{d}{{dx}}\left( {x + 5} \right) = \dfrac{{dt}}{{dx}}$
$ \Rightarrow dx = dt$
Now change the integral limits.
So when, x= -4, so from equation (3) we have,
$ \Rightarrow - 4 + 5 = t$
$ \Rightarrow t = 1$
Now when, x = -5, so from equation (1) we have,
$ \Rightarrow - 5 + 5 = t$
$ \Rightarrow t = 0$
So integral limits is changed from (1 to 0)
Now substitute these values in equation (2) we have,
 $ \Rightarrow {I_1} = \int_{ - 4}^{ - 5} {{e^{{{\left( {x + 5} \right)}^2}}}dx} = \int_1^0 {{e^{{t^2}}}dt} $............... (4)
Now simplify integral ${I_2}$ we have,
${I_2} = \int_{\dfrac{1}{3}}^{\dfrac{2}{3}} {{e^{9{{\left( {x - \dfrac{2}{3}} \right)}^2}}}dx} $................. (5)
In the above integral substitute ($x - \dfrac{2}{3} = \dfrac{t}{3}$)............ (6)
Now differentiate equation (6) w.r.t x we have,
$ \Rightarrow \dfrac{d}{{dx}}\left( {x - \dfrac{2}{3}} \right) = \dfrac{d}{{dx}}\left( {\dfrac{t}{3}} \right)$
$ \Rightarrow 3dx = dt$
Now change the integral limits.
So when, x = (1/3), so from equation (6) we have,
$ \Rightarrow \dfrac{1}{3} - \dfrac{2}{3} = \dfrac{t}{3}$
$ \Rightarrow t = - 1$
Now when, x = (2/3), so from equation (6) we have,
$ \Rightarrow \dfrac{2}{3} - \dfrac{2}{3} = \dfrac{t}{3}$
$ \Rightarrow t = 0$
So integral limits is changed from (-1 to 0)
Now substitute these values in equation (5) we have,
 $ \Rightarrow {I_2} = \int_{\dfrac{1}{3}}^{\dfrac{2}{3}} {{e^{9{{\left( {x - \dfrac{2}{3}} \right)}^2}}}dx} = \int_{ - 1}^0 {{e^{9{{\left( {\dfrac{t}{3}} \right)}^2}}}dt} = \int_{ - 1}^0 {{e^{{t^2}}}dt} $............... (7)
Now substitute the value from equations (4) and (7) in equation (1) we have,
$ \Rightarrow I = \int_1^0 {{e^{{t^2}}}dt} + \int_{ - 1}^0 {{e^{{t^2}}}dt} $
Now according to definite integral property if function is even than $\int_{ - a}^0 {f\left( x \right)dx} = - \int_a^0 {f\left( x \right)dx} $ so use this property in the above equation we have,
As we all know that for function to be even f (-x) = f (x)
So in the above integral, $f\left( t \right) = {e^{{t^2}}}$
Therefore, $f\left( { - t} \right) = {e^{{{\left( { - t} \right)}^2}}} = {e^{{t^2}}}$
So the integral function is even.
$ \Rightarrow I = \int_1^0 {{e^{{t^2}}}dt} + \int_{ - 1}^0 {{e^{{t^2}}}dt} = \int_1^0 {{e^{{t^2}}}dt} - \int_1^0 {{e^{{t^2}}}dt} $
$ \Rightarrow I = \int_1^0 {{e^{{t^2}}}dt} - \int_1^0 {{e^{{t^2}}}dt} = 0$
So this is the required answer.
Hence option (C) is the correct answer.

Note:Whenever we face such types of questions the key concept we have to remember is that according to definite integral property, $\int_{ - a}^0 {f\left( x \right)dx} = - \int_a^0 {f\left( x \right)dx} $ if and only if the function f (x) is even, and the condition of even function is, f (-x) = f (x).so first simplify the integral using integration by substitution as above then applied above described properties we will get the required answer.