
The value of \[\int_{ - 5}^5 {\log \left( {\dfrac{{130 - {x^3}}}{{130 + {x^3}}}} \right)} dx\] is equal to
A. $\log \dfrac{{57}}{5}$
B. \[2\int_{ - 5}^5 {\log \left( {\dfrac{{130 - {x^3}}}{{130 + {x^3}}}} \right)} dx\]
C. 0
D. -1
Answer
600k+ views
Hint: We will use the property $\int_{ - a}^a {f(x)dx = 2\int_0^a {f(x)} } $ when f(x) is even and $\int_{ - a}^a {f(x)dx = 0} $ when f(x) is odd. We will also use the property of log functions which is $\log \left( {\dfrac{m}{n}} \right) = \log m - \log n$.
Complete step by step solution:
Now, we are given $I = \int_{ - 5}^5 {\log \left( {\dfrac{{130 - {x^3}}}{{130 + {x^3}}}} \right)} dx$. We will use the property $\int_{ - a}^a {f(x)dx = 2\int_0^a {f(x)} } $ and $\int_{ - a}^a {f(x)dx = 0} $ when f(x) is even and odd respectively.
Now, a function is even when f(x) = f(-x). and the function is odd when f(x) = -f(-x) We will check whether the function is even and odd by replacing x by -x in f(x).
Now, in the given question, f(x) = $\log \left( {\dfrac{{130 - {x^3}}}{{130 + {x^3}}}} \right)$. Replacing x by -x, we get
f(-x) = $\log \left( {\dfrac{{130 - {{( - x)}^3}}}{{130 + {{( - x)}^3}}}} \right)$ = $\log \left( {\dfrac{{130 + {x^3}}}{{130 - {x^3}}}} \right)$
Now, from the property of log functions, we know that $\log \left( {\dfrac{m}{n}} \right) = \log m - \log n$
So, we can write f(-x) = $\log (130 + {x^3}) - \log (130 - {x^3})$
By using the above property, we can say f(x) = $\log (130 - {x^3}) - \log (130 + {x^3})$
Therefore, we get
f(x) = -f(-x) so, f(x) is an odd function.
Now, we know that $\int_{ - a}^a {f(x)} = 0$ when f(x) is an odd function. So, we get
$I = \int_{ - 5}^5 {\log \left( {\dfrac{{130 - {x^3}}}{{130 + {x^3}}}} \right)} $ = 0
So, \[\int_{ - 5}^5 {\log \left( {\dfrac{{130 - {x^3}}}{{130 + {x^3}}}} \right)} dx\] = 0
So, option (C) is correct.
Note: Whenever we come up with such types of questions, we will use the properties of integration. Various properties of integration help us in solving difficult integration. We use such properties in this question because we don’t know the integration of log x, if we try to solve this question through integration by – parts method, we will not be able to solve the given question.
Complete step by step solution:
Now, we are given $I = \int_{ - 5}^5 {\log \left( {\dfrac{{130 - {x^3}}}{{130 + {x^3}}}} \right)} dx$. We will use the property $\int_{ - a}^a {f(x)dx = 2\int_0^a {f(x)} } $ and $\int_{ - a}^a {f(x)dx = 0} $ when f(x) is even and odd respectively.
Now, a function is even when f(x) = f(-x). and the function is odd when f(x) = -f(-x) We will check whether the function is even and odd by replacing x by -x in f(x).
Now, in the given question, f(x) = $\log \left( {\dfrac{{130 - {x^3}}}{{130 + {x^3}}}} \right)$. Replacing x by -x, we get
f(-x) = $\log \left( {\dfrac{{130 - {{( - x)}^3}}}{{130 + {{( - x)}^3}}}} \right)$ = $\log \left( {\dfrac{{130 + {x^3}}}{{130 - {x^3}}}} \right)$
Now, from the property of log functions, we know that $\log \left( {\dfrac{m}{n}} \right) = \log m - \log n$
So, we can write f(-x) = $\log (130 + {x^3}) - \log (130 - {x^3})$
By using the above property, we can say f(x) = $\log (130 - {x^3}) - \log (130 + {x^3})$
Therefore, we get
f(x) = -f(-x) so, f(x) is an odd function.
Now, we know that $\int_{ - a}^a {f(x)} = 0$ when f(x) is an odd function. So, we get
$I = \int_{ - 5}^5 {\log \left( {\dfrac{{130 - {x^3}}}{{130 + {x^3}}}} \right)} $ = 0
So, \[\int_{ - 5}^5 {\log \left( {\dfrac{{130 - {x^3}}}{{130 + {x^3}}}} \right)} dx\] = 0
So, option (C) is correct.
Note: Whenever we come up with such types of questions, we will use the properties of integration. Various properties of integration help us in solving difficult integration. We use such properties in this question because we don’t know the integration of log x, if we try to solve this question through integration by – parts method, we will not be able to solve the given question.
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