Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
If I=1+x41-x432dx=1fx+C (where, C is the constant of integration) and f2=-154, then the value of 2f12 is
MathematicsIndefinite IntegrationJEE Main
Solution:
2717 Upvotes Verified Answer
The correct answer is: 3
Given integral is
I=1x3+x1x2-x232dx
Let 1x2-x2=t
-2x3-2xdx=dt
I=-12dtt32
=t-12+C
=11x2-x2+C
fx=1x2-x2
f12=2-12=32
2f12=3

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.