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The area bounded by $y equals sin to the power of negative 1 end exponent invisible function application blank x equals fraction numerator 1 over denominator square root of 2 end fraction$ and $x$ -axis is

$(\frac{1}{\sqrt{2}} + 1)$ sq unit
$(1 - \frac{1}{\sqrt{2}})$ sq unit
$\frac{π}{4 \sqrt{2}} s q u n i t$
$(\frac{π}{4 \sqrt{2}} + \frac{1}{\sqrt{2}} - 1)$ sq unit

hint Hint:

In this question we have to $y equals sin to the power of negative 1 end exponent open parentheses x close parentheses$ and the line is $x equals fraction numerator 1 over denominator square root of 2 end fraction$ . We have to find the area bounded inside the point of intersection of these two curves and above the x-axis. We have to use ILATE to integrate $I equals space u cross times integral v d x space minus space integral open parentheses fraction numerator d u over denominator d x end fraction cross times integral v d x close parentheses d x$ here $u equals 1 comma space v space equals space sin to the power of negative 1 end exponent open parentheses x close parentheses$ . After integrating we put the limit $x equals 0 space t o space x equals fraction numerator 1 over denominator square root of 2 end fraction$

The correct answer is: $open parentheses fraction numerator pi over denominator 4 square root of 2 end fraction plus fraction numerator 1 over denominator square root of 2 end fraction minus 1 close parentheses blank$ sq unit

In This figure the green curve is $y equals sin to the power of negative 1 end exponent open parentheses x close parentheses$ and the red line is $x equals fraction numerator 1 over denominator square root of 2 end fraction$
We have to find the area bounded inside the point of intersection of these two curves and above the x-axis
The point of intersection is $open parentheses fraction numerator 1 over denominator square root of 2 end fraction comma straight pi over 4 close parentheses$
The area will be equal to $integral subscript 0 superscript fraction numerator 1 over denominator square root of 2 end fraction end superscript sin to the power of negative 1 end exponent open parentheses x close parentheses d x$
We know that for integration of $sin to the power of negative 1 end exponent open parentheses x close parentheses$ we will use ILATE rule
$I space equals space integral 1 cross times sin to the power of negative 1 end exponent open parentheses x close parentheses$ let $u equals 1 comma space v space equals space sin to the power of negative 1 end exponent open parentheses x close parentheses$
$I equals space u cross times integral v d x space minus space integral open parentheses fraction numerator d u over denominator d x end fraction cross times integral v d x close parentheses d x$
By substituting the values of $u comma v$ in the above equation we get
$I space equals space x sin to the power of negative 1 end exponent open parentheses x close parentheses minus integral fraction numerator x over denominator square root of 1 minus x squared end root end fraction d x$
Let $1 minus x squared equals t squared$
=> By differentiating both sides we get,
=> $negative x d x equals t d t$
=> $I space equals space x sin to the power of negative 1 end exponent open parentheses x close parentheses minus integral t over t d t$
=> $I space equals space x sin to the power of negative 1 end exponent open parentheses x close parentheses minus t$
=> $I space equals space x sin to the power of negative 1 end exponent open parentheses x close parentheses minus square root of 1 minus x squared end root space plus c$ where c is the constant of Integration.
By putting the limits we get
=> $I space equals open square brackets space x sin to the power of negative 1 end exponent open parentheses x close parentheses minus square root of 1 minus x squared end root space plus c close square brackets subscript 0 superscript fraction numerator 1 over denominator square root of 2 end fraction end superscript$
=> $open parentheses fraction numerator straight pi over denominator 4 square root of 2 end fraction plus fraction numerator 1 over denominator square root of 2 end fraction minus 1 close parentheses s q. space u n i t$

Simplify : 2m-3 =17

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The area bounded by and -axis is

sq unit sq unit sq unit

In this question we have to and the line is . We have to find the area bounded inside the point of intersection of these two curves and above the x-axis. We have to use ILATE to integrate here . After integrating we put the limit

The correct answer is: sq unit

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The area bounded by $y equals sin to the power of negative 1 end exponent invisible function application blank x equals fraction numerator 1 over denominator square root of 2 end fraction$ and $x$ -axis is

$(\frac{1}{\sqrt{2}} + 1)$ sq unit
$(1 - \frac{1}{\sqrt{2}})$ sq unit
$\frac{π}{4 \sqrt{2}} s q u n i t$
$(\frac{π}{4 \sqrt{2}} + \frac{1}{\sqrt{2}} - 1)$ sq unit

The correct answer is: $open parentheses fraction numerator pi over denominator 4 square root of 2 end fraction plus fraction numerator 1 over denominator square root of 2 end fraction minus 1 close parentheses blank$ sq unit

find the value of m in $fraction numerator 9 over denominator text m end text end fraction equals 3$

find the value of m in $fraction numerator 9 over denominator text m end text end fraction equals 3$

simplify $equals fraction numerator a squared cross times 3 to the power of 4 over denominator a 3 squared end fraction$

simplify $equals fraction numerator a squared cross times 3 to the power of 4 over denominator a 3 squared end fraction$

Represent $36 over 169$ in powers of prime numbers

Represent $36 over 169$ in powers of prime numbers

what is $40 straight percent sign$ of 1080

what is $40 straight percent sign$ of 1080