Question

The value of $\int_{ - 10}^{10} {\dfrac{{dx}}{{{e^{{x^5}}} + 1}}}$ is equal to ?

Answer 1

Hint : The given question requires us to compute the value of an integral given to us in the question. We are also given limits for the integral that means that we have to compute a definite integral with upper and lower limits. In order to solve such definite integrals, we are required to first compute the indefinite integral and then substitute the upper and lower limits and find the required numeric value or we can also make use of some definite integral properties to make the integral easier and less time consuming.

Complete step by step solution:
So, we have to find the value of $\int_{ - 10}^{10} {\dfrac{{dx}}{{{e^{{x^5}}} + 1}}}$ .
We can either find the indefinite integral of the expression and then put in the limits, or we can make use of some of the definite integral properties. On observing the limits of the definite integral $\int_{ - 10}^{10} {\dfrac{{dx}}{{{e^{{x^5}}} + 1}}}$ given to us, we notice that the limits of the integral are negative of each other.
We know that the definite integral of the form $\int_a^b {f\left( x \right)} \,dx$ can be computed easily using a definite integral property. According to the property, the definite integral $\int_a^b {f\left( x \right)} \,dx = \int_a^b {f\left( {a + b - x} \right)} \,dx$ .
So, $I = \int_{ - 10}^{10} {\dfrac{1}{{{e^{{x^5}}} + 1}}} \,dx - - - - - \left( 1 \right)$
Using the definite integral property $\int_a^b {f\left( x \right)} \,dx = \int_a^b {f\left( {a + b - x} \right)} \,dx$ , we get,
$\Rightarrow \int_{ - 10}^{10} {\dfrac{1}{{{e^{{{\left( { - 10 + 10 - x} \right)}^5}}} + 1}}} \,dx$
Cancelling the like terms with opposite signs, we get,
$\Rightarrow \int_{ - 10}^{10} {\dfrac{1}{{{e^{{{\left( { - x} \right)}^5}}} + 1}}} \,dx$
$\Rightarrow \int_{ - 10}^{10} {\dfrac{1}{{{e^{ - {x^5}}} + 1}}} \,dx$
Simplifying further,
$\Rightarrow \int_{ - 10}^{10} {\dfrac{1}{{\dfrac{1}{{{e^{{x^5}}}}} + 1}}} \,dx$
$\Rightarrow \int_{ - 10}^{10} {\dfrac{1}{{\dfrac{{1 + {e^{{x^5}}}}}{{{e^{{x^5}}}}}}}} \,dx$
$\Rightarrow \int_{ - 10}^{10} {\dfrac{{{e^{{x^5}}}}}{{1 + {e^{{x^5}}}}}} \,dx - - - - - \left( 2 \right)$
Adding $\left( 1 \right)$ and $\left( 2 \right)$ , we get,
$2I = \int_{ - 10}^{10} {\dfrac{1}{{{e^{{x^5}}} + 1}}} \,dx + \int_{ - 10}^{10} {\dfrac{{{e^{{x^5}}}}}{{1 + {e^{{x^5}}}}}} \,dx$
Combining both the integrals, we get,
$\Rightarrow 2I = \int_{ - 10}^{10} {\dfrac{1}{{{e^{{x^5}}} + 1}}} \, + \dfrac{{{e^{{x^5}}}}}{{1 + {e^{{x^5}}}}}dx$
$\Rightarrow 2I = \int_{ - 10}^{10} {\dfrac{{1 + {e^{{x^5}}}}}{{1 + {e^{{x^5}}}}}} \,dx$
Cancelling the numerator and denominator, we get,
$\Rightarrow 2I = \int_{ - 10}^{10} 1 \,dx$
Putting the upper and lower limits of integral, we get,
$\Rightarrow 2I = \left( {10} \right) - \left( { - 10} \right)$
$\Rightarrow 2I = 20$
Dividing both sides of the equation by $2$ , we get,
$\Rightarrow I = 10$
The value of $\int_{ - 10}^{10} {\dfrac{{dx}}{{{e^{{x^5}}} + 1}}}$ is equal to $10$ .
So, the correct answer is “10 ”.

Note : We should learn properties related to definite integral before solving integrals as one given in the question itself. Such properties are of great use when it comes evaluating complex definite integrals with some conditions on the limits.