The value of integral $\int\limits_4^0 {\dfrac{{[{x^2}]dx}}{{[{x^2} - 28x + 196] + [{x^2}]}},where\,[x]} $ denotes the greatest integer less than or equal to x, is :
A) $1/3$
B) $6$
C) $7$
D) $3$
Answer
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Hint:Use the property of integral $\int\limits_a^b {f(x)} dx = \int\limits_a^b {f(a + b - x)dx} $ to obtain a new integral. Then add the two integrals and get the value for the original integral.
Complete step-by-step answer:
Let us assume that
$I = \int\limits_4^{10} {\dfrac{{[{x^2}]dx}}{{[{x^2} - 28x + 196] + [{x^2}]}}} $
If you carefully observe , you will realize that ${x^2} - 28x + 196 = {(x - 14)^2}$ as ${(a - b)^2} = {a^2} - 2ab + {b^2}$
$ = > I = \int\limits_4^{10} {\dfrac{{[{x^2}]}}{{{{[(x - 14)]}^2} + [{x^2}]}}} ..................(1)$
Now we will use the property
$\int\limits_a^b {f(x)} dx = \int\limits_a^b {f(a + b - x)dx} $
Hence we have to replace $x \,as\, (a + b - x)$ that is in this case $x \,as\, > (14 - x)$
$I = \int\limits_4^{10}
\dfrac{{[14 - {x^2}]}}{{[{{(14 - x - 14)}^2}] + [{{(14 - x)}^2}]}} \\
\\ \\
= > \int\limits_4^{10} {\dfrac{{{{[14 - x]}^2}}}{{{{[x]}^2} + {{[(14 - x)]}^2}}}.........................(2)} \\ $
Adding both the equation we get
$ I + I = \int\limits_4^{10} {\dfrac{{[{x^2}]dx}}{{[{x^2}] + {{[x - 14]}^2}}} + \int\limits_4^{10} {\dfrac{{{{[14 - x]}^2}dx}}{{{{[14 - x]}^2} + [{x^2}]}}} } \\
\text{since},{(a - b)^2} = {(b - a)^2} \\
{(x - 14)^2} = {(14 - x)^2} \\
2I = \int\limits_4^{10} {\dfrac{{[{x^2}]dx}}{{[{x^2}] + {{[x - 14]}^2}}} + \int\limits_4^{10} {\dfrac{{{{[x - 14]}^2}dx}}{{{{[x - 14]}^2} + [{x^2}]}}} } \\
2I = \int\limits_4^{10} {\dfrac{{[{x^2}] + {{[(x - 14)]}^2}}}{{[{x^2}] + {{[x - 14]}^2}}}dx} \\
2I = \int\limits_4^{10} {dx} \\
2I = |x|_4^{10} \\
2I = 10 - 4 = 6 \\
I = 3 \\ $
Hence the value of integral is 3
So, the correct answer is “Option D”.
Note:The property we used in the question is quite important. This is because without using this property, this question is too difficult to solve. Hence, we need to take care of this property whenever we ask such questions. It is generally used in questions where the denominator has 2 or more terms in addition and the numerator has a term common with the denominator such as we used in the above question.
Complete step-by-step answer:
Let us assume that
$I = \int\limits_4^{10} {\dfrac{{[{x^2}]dx}}{{[{x^2} - 28x + 196] + [{x^2}]}}} $
If you carefully observe , you will realize that ${x^2} - 28x + 196 = {(x - 14)^2}$ as ${(a - b)^2} = {a^2} - 2ab + {b^2}$
$ = > I = \int\limits_4^{10} {\dfrac{{[{x^2}]}}{{{{[(x - 14)]}^2} + [{x^2}]}}} ..................(1)$
Now we will use the property
$\int\limits_a^b {f(x)} dx = \int\limits_a^b {f(a + b - x)dx} $
Hence we have to replace $x \,as\, (a + b - x)$ that is in this case $x \,as\, > (14 - x)$
$I = \int\limits_4^{10}
\dfrac{{[14 - {x^2}]}}{{[{{(14 - x - 14)}^2}] + [{{(14 - x)}^2}]}} \\
\\ \\
= > \int\limits_4^{10} {\dfrac{{{{[14 - x]}^2}}}{{{{[x]}^2} + {{[(14 - x)]}^2}}}.........................(2)} \\ $
Adding both the equation we get
$ I + I = \int\limits_4^{10} {\dfrac{{[{x^2}]dx}}{{[{x^2}] + {{[x - 14]}^2}}} + \int\limits_4^{10} {\dfrac{{{{[14 - x]}^2}dx}}{{{{[14 - x]}^2} + [{x^2}]}}} } \\
\text{since},{(a - b)^2} = {(b - a)^2} \\
{(x - 14)^2} = {(14 - x)^2} \\
2I = \int\limits_4^{10} {\dfrac{{[{x^2}]dx}}{{[{x^2}] + {{[x - 14]}^2}}} + \int\limits_4^{10} {\dfrac{{{{[x - 14]}^2}dx}}{{{{[x - 14]}^2} + [{x^2}]}}} } \\
2I = \int\limits_4^{10} {\dfrac{{[{x^2}] + {{[(x - 14)]}^2}}}{{[{x^2}] + {{[x - 14]}^2}}}dx} \\
2I = \int\limits_4^{10} {dx} \\
2I = |x|_4^{10} \\
2I = 10 - 4 = 6 \\
I = 3 \\ $
Hence the value of integral is 3
So, the correct answer is “Option D”.
Note:The property we used in the question is quite important. This is because without using this property, this question is too difficult to solve. Hence, we need to take care of this property whenever we ask such questions. It is generally used in questions where the denominator has 2 or more terms in addition and the numerator has a term common with the denominator such as we used in the above question.
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