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If ƒ(x) = `a + bx + cx^2`, whre c > 0 and `b^2` – 4ac < 0, then the area enclosed by the coordinate axes, the line x = 2 and the curve y = ƒ(x) is given by :
A. |
`1/3{4f(1) + f(2)}` |
B. |
`1/2 { f(0)+4f(1)+f(2)}` |
C. |
`1/2 {f(0)+4f(1)}` |
D. |
`1/3 { f(0)+4 f (1)+ f (2)}` |
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Area =`int_0^2(a+bxc+cx^2)dx`
`= 2a+2b+8/3c`
= `1/3 [6a 6b 8c]` ..........(i)
But, ƒ(x) = a + bx + `cx^3`
ƒ(0) = a,
ƒ(1) = a + b + c
`ƒ(2) = a + 2b + 4c`
`1/3{ƒ(0)+4ƒ(1)+ƒ(2)}`
`1/3 {a+4(a+b+c)+(a+2b+4)}`
`1/3(6a+6b+8c)`