The value of -5-5fxdx-k- where fx-minx-1-x-1- x - and - denotes fractional part of x- Then the value of k is equal to

We know, {x+1}={x 1}={x} Thus, f(x)=min({x+1},{x 1})={x} ⇒ nbsp; I=∫^ 5_5f(x) dx ⇒ I nbsp;=∫^5_ 5{x}dx ⇒ I nbsp;=[5 ( 5)]∫^1_0{x}dx [∵{x} is periodic with ...

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Standard XII

Mathematics

Integration Using Substitution

The value of ...

Question

The value of $5 \int - 5 f (x) d x = k,$ where $f (x) = min ({x + 1}, {x - 1}), \forall x \in R$ and ${\cdot}$ denotes fractional part of $x$ . Then the value of $k$ is equal to

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Solution

We know,
${x + 1} = {x - 1} = {x}$
Thus,
$f (x) = min ({x + 1}, {x - 1}) = {x}$
$\Rightarrow I = - 5 \int 5 f (x) d x \Rightarrow I = 5 \int - 5 {x} d x \Rightarrow I = [5 - (- 5)] 1 \int 0 {x} d x [∵ {x} is periodic with period 1] ∴ I = 10 1 \int 0 x d x = 5$

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Integration Using Substitution

Standard XII Mathematics

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The value of 5∫−5f(x)dx=k, where f(x)=min({x+1},{x−1}),∀x∈R and {⋅} denotes fractional part of x. Then the value of k is equal to

We know, {x+1}={x−1}={x} Thus, f(x)=min({x+1},{x−1})={x} ⇒I=−5∫5f(x) dx⇒I=5∫−5{x}dx⇒I=[5−(−5)]1∫0{x}dx[∵{x} is periodic with period 1]∴I=101∫0x dx=5

The value of $5 \int - 5 f (x) d x = k,$ where $f (x) = min ({x + 1}, {x - 1}), \forall x \in R$ and ${\cdot}$ denotes fractional part of $x$ . Then the value of $k$ is equal to

We know,
${x + 1} = {x - 1} = {x}$
Thus,
$f (x) = min ({x + 1}, {x - 1}) = {x}$
$\Rightarrow I = - 5 \int 5 f (x) d x \Rightarrow I = 5 \int - 5 {x} d x \Rightarrow I = [5 - (- 5)] 1 \int 0 {x} d x [∵ {x} is periodic with period 1] ∴ I = 10 1 \int 0 x d x = 5$