+91-8076753736[email protected]
logo
logo
-->

Vector Algebra

Question
CBSEENMA12032815
Wired Faculty App

Area bounded by the curve y = x3  the x-axis and the ordinates x - 2 and a = 1 is

  • -9

  • fraction numerator negative 15 over denominator 4 end fraction
  • 15 over 4
  • 17 over 4

Solution

D.

17 over 4

We want to find the area bounded by the curve y = x3, the x-axis and the ordinates x = - 2 and x = 1.
A rough sketch of the curve y = x3 is given in the figure.

Required area  = integral subscript negative 2 end subscript superscript 1 space open vertical bar straight x cubed close vertical bar space dx space equals space integral subscript negative 2 end subscript superscript 0 open vertical bar straight x cubed close vertical bar space dx space plus space integral subscript 0 superscript 1 open vertical bar straight x close vertical bar cubed space dx space equals space integral subscript negative 2 end subscript superscript 0 left parenthesis negative straight x cubed right parenthesis space dx plus integral subscript 0 superscript 1 straight x cubed space dx
                         equals negative open square brackets straight x to the power of 4 over 4 close square brackets subscript 2 superscript 0 plus open square brackets straight x to the power of 4 over 4 close square brackets subscript 0 space equals space minus open square brackets 0 minus fraction numerator left parenthesis negative 2 right parenthesis to the power of 4 over denominator 4 end fraction close square brackets plus open square brackets 1 fourth minus 0 close square brackets space equals space 4 plus 1 fourth space equals space 17 over 4
therefore space space space left parenthesis straight D right parenthesis space is space correct space answer.

Some More Questions From Vector Algebra Chapter

Mock Test Series

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation