Question

The sum of the integrals \[\int_{ - 4}^{ - 5} {{e^{{{(x + 5)}^2}}}dx} + \int_{\dfrac{1}{3}}^{\dfrac{2}{3}} {{e^{9{{(x - \dfrac{2}{3})}^2}}}dx} \] is
A. \[{e^5}\]
B. \[{e^4}\]
C. \[3{e^2}\]
D. \[0\]

Answer 1

Hint: The given question deals the concept of definite integration. A definite integral in mathematics, \[\int_a^b {f(x)dx} \]Is the area of the x y-plane bounded by the graph f, the x-axis, and the lines x = a and x = b, so that the area above the x-axis adds to the total and the area below the x-axis subtracts from the total. To solve the given question, we will solve the given integrals in parts and find the solution.

Complete step by step solution:
Let the given equation be,
\[I = \int_{ - 4}^{ - 5} {{e^{{{(x + 5)}^2}}}dx} + \int_{\dfrac{1}{3}}^{\dfrac{2}{3}} {{e^{9{{(x - \dfrac{2}{3})}^2}}}dx} \]
Here, we will solve the given question in two parts.
So let the first part be,
\[{I_1} = \int_{ - 4}^{ - 5} {{e^{{{(x + 5)}^2}}}dx} \]
And the second part be,
\[{I_2} = \int_{\dfrac{1}{3}}^{\dfrac{2}{3}} {{e^{9{{(x - \dfrac{2}{3})}^2}}}dx} \]

Firstly, we will solve the first part,
Therefore, we have,
\[ \Rightarrow {I_1} = \int_{ - 4}^{ - 5} {{e^{{{(x + 5)}^2}}}dx} \]
Let,
Here, let \[x + 5 = t - - - - - (1)\]
Therefore, \[dx = dt\]
Now, using the lower limit in equation (1) when we put , \[x = - 4\]
Therefore, we get \[t = 1\]
Using the upper limit, \[x = - 5\]and this gives\[t = 0\].
Therefore, we get \[{I_1} = \int_1^0 {{e^{{t^2}}}dt} = \int_1^0 {{e^{{x^2}}}dx} - - - (2)\](after substituting dt into dx)
Now, for the second part, we have,
\[ \Rightarrow {I_2} = 3\int_{\dfrac{1}{3}}^{\dfrac{2}{3}} {{e^{9{{(x - \dfrac{2}{3})}^2}}}dx} \]

Let, \[3x - 2 = - u - - - - - (3)\]
Therefore, \[3dx = - du\]
Using lower limit in (3) we put,
\[x = \dfrac{1}{3}\], we get \[u = 1\]
And, using upper limit in (3)
We have, \[x = \dfrac{2}{3}\]therefore we get, \[u = 0\]
Thus we get,
\[ \Rightarrow {I_2} = - \int_1^0 {{e^{{u^2}}}du} = - \int_1^0 {{e^{{x^2}}}dx} \]
Now, we add \[{I_1} + {I_2}\]to find the ultimate solution,
\[ \therefore {I_1} + {I_2} = \int_1^0 {{e^{{x^2}}}dx} - \int_1^0 {{e^{{x^2}}}dx = 0} \]

Thus, the correct option is option D.

Note: The important thing to note here is the difference between definite and indefinite integrals. In an indefinite integral, there are no limits to integration. A definite integral is a number when the lower and upper limits are constants. The indefinite integral is a family of functions whose derivatives are f. The difference between the two functions in the family is a constant. If the definite integral is evaluated by first computing an indefinite integral and then replacing the integration boundary with the result, we must be aware that indefinite integration could lead to discontinuities.