Let a and b be real constants such that the function f defined by f (x )=(x^2+3x+a ,x 1 bx+2 ,x1) be differentiable on R. Then, the value of _-2^2 0.05emf (x ) dx equals

Question

Let a and b be real constants such that the function f defined by f (x )=(x^2+3x+a ,x 1 bx+2 ,x>1) be differentiable on R. Then, the value of _-2^2 0.05emf (x ) dx equals

Accepted Answer

Correct answer: D. (d) f is continuous f^ ' (x )=2x+3 ,k<1 ∴ 4+a=b+2 a=b-2 b ,x>1 a=b-2 f is differentiable ∴ b=5 ∴ a=3 0.05em 0.05em(x^2+3x+3) dx+ 0.05em (5x+2 ) dx =( x^33+ 3x^22+3x)_-2^1+( 5x^22+2x)_1^2 =( 13+ 32+3)-( -83+6-6)+(10+4- 52-2) =6+ 32+12- 52=17

$Let a and b be real constants such that the function f defined by f (x) = (x^{2} + 3 x + a, x \leq 1 bx + 2, x > 1) be differentiable on R . Then, the value of \int_{- 2}^{2} f (x) dx equals$

Solution

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