The value of the integral 10∫4[x2]dx[x2−28x+196]+[x2],where[x] denotes the greatest integer less than or equal to x, is :
A
3
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
7
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
13
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
6
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is A3 I=10∫4[x2]dx[x2−28x+196]+[x2]=10∫4[x2]dx[(14−x)2]+[x2]Usingb∫af(a+b−x)=b∫af(x)I=10∫4[(14−x)2]dx[(14−x)2]+[x2]⇒2I=10∫4dx⇒2I=10−4⇒I=3