Definite Integration Question 35

Question 35

  1. If f(x)=Asinπx2+B,

f12=2 and 01f(x)dx=2Aπ, then constants

A and B are

(1995, 2M) (a) π2 and π2 (b) 2π and 3π (c) 0 and 4π (d) 4π and 0

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Solution:

  1. Given, f(x)=Asinπx2+B,f12=2

 and 01f(x)dx=2Aπ

f(x)=Aπ2cosπx2f12=Aπ2cosπ4=Aπ22

 But f12=2Aπ22=2A=4π

Now, 01f(x)dx=2Aπ01Asinπx2+Bdx=2Aπ

2Aπcosπx2+Bx=2AπB+2Aπ=2Aπ B=0



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