#### Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Click to Chat

1800-1023-196

+91 7353221155

CART 0

• 0
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

# integral of  [f(x)] between the limits 2 to 6 , wheref(x)=[(x-1)^2] /2[x]+1 , where [.] represents greatest integer function .

Jitender Singh IIT Delhi
7 years ago
Ans:
$I = \int_{2}^{6}[\frac{[(x-1)^{2}]}{2[x]+1}]dx$
$2\leq x< 3,[\frac{[(x-1)^{2}]}{2[x]+1}]=0$
$3\leq x< \sqrt{7}+1,[\frac{[(x-1)^{2}]}{2[x]+1}]=0$
$\sqrt{7}+1\leq x< 4,[\frac{[(x-1)^{2}]}{2[x]+1}]=1$
$4\leq x< 5,[\frac{[(x-1)^{2}]}{2[x]+1}]=1$
$5\leq x< \sqrt{22}+1,[\frac{[(x-1)^{2}]}{2[x]+1}]=1$
$\sqrt{22}+1\leq x< 6, [\frac{[(x-1)^{2}]}{2[x]+1}]=2$
$I = \int_{2}^{\sqrt{7}+1}[f(x)]dx+\int_{\sqrt{7}+1}^{\sqrt{22}+1}[f(x)]dx+\int_{\sqrt{22}+1}^{6}[f(x)]dx$
$I = \int_{2}^{\sqrt{7}+1}0.dx+\int_{\sqrt{7}+1}^{\sqrt{22}+1}1.dx+\int_{\sqrt{22}+1}^{6}2.dx$
$I = (x)_{\sqrt{7}+1}^{\sqrt{22}+1}+(2x)_{\sqrt{22}+1}^{6}$
$I = \sqrt{22}-\sqrt{7}+10-2.\sqrt{22}$
$I = 10-\sqrt{22}-\sqrt{7}$
Thanks & Regards
Jitender Singh
IIT Delhi