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Maths-

General

Easy

Question

The area bounded by mini $left parenthesis vertical line x vertical line comma vertical line y vertical line right parenthesis equals 2 text end text$ and $m a x left parenthesis vertical line x vertical line comma vertical line y vertical line right parenthesis equals 4 text is : end text$

4
8
16
32

The correct answer is: 16

Required are = 4[2 ´ 2] = 16

Related Questions to study

$f left parenthesis x right parenthesis equals stretchy integral subscript 1 end subscript superscript x end superscript fraction numerator tan to the power of negative 1 end exponent invisible function application t over denominator t end fraction d t semicolon x greater than 0$ then the value of $f open parentheses e to the power of 2 end exponent close parentheses minus f open parentheses fraction numerator 1 over denominator e to the power of 2 end exponent end fraction close parentheses text is end text$

$f left parenthesis x right parenthesis equals stretchy integral subscript 1 end subscript superscript x end superscript fraction numerator tan to the power of negative 1 end exponent invisible function application t over denominator t end fraction d t semicolon x greater than 0$ then the value of $f open parentheses e to the power of 2 end exponent close parentheses minus f open parentheses fraction numerator 1 over denominator e to the power of 2 end exponent end fraction close parentheses text is end text$

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If $text end text g left parenthesis x right parenthesis equals stretchy integral subscript 0 end subscript superscript x end superscript left parenthesis vertical line s i n invisible function application t vertical line plus vertical line c o s invisible function application t vertical line right parenthesis d t text then end text g open parentheses x plus fraction numerator n pi over denominator 2 end fraction close parentheses$ is equal to where $n element of N$

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parallel

If the primitive of $text end text s i n to the power of negative 3 divided by 2 end exponent invisible function application x s i n to the power of negative 1 divided by 2 end exponent invisible function application left parenthesis x plus theta right parenthesis text end text text is end text text end text minus 2 c o s e c invisible function application theta square root of f left parenthesis x right parenthesis end root plus C text end text$ then

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