Let f : R → R be a differentiable function having f (2) = 6, f′ (2) =(1/48) . Then $limit as straight x space rightwards arrow 2 of space integral subscript 6 superscript straight f left parenthesis straight x right parenthesis end superscript space fraction numerator 4 straight t cubed over denominator straight x minus 2 end fraction dt space equals$

24

36

12

18

Name: Wired Faculty - The Best Learning App
Rating: 4.6 (1864 reviews)

Solution

limit as straight x space rightwards arrow 2 of space integral subscript 0 superscript straight f left parenthesis straight x right parenthesis end superscript fraction numerator 4 straight t cubed over denominator straight x minus 2 end fraction dt Applying space straight L space hospital space rule limit as straight x rightwards arrow 2 of space open square brackets 4 straight f space left parenthesis straight x right parenthesis squared space straight f apostrophe space left parenthesis straight x right parenthesis close square brackets space equals space 4 straight f space left parenthesis 2 right parenthesis cubed space straight f apostrophe space left parenthesis 2 right parenthesis space equals space 4 space straight x space 6 cubed space straight x 1 over 48 space equals space 18

For Daily Free Study Material Join wiredfaculty