**Let f(x)= 4-3x. Find K so that the integral from -1 to x of f(t)dt+k = the integral from 3 to x of f(t)?**

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justinbains132,

the intergral from -1 to x of f(t)dt+k is

(-3/2)*x^2 + (4 + k)*x + 11/2 + k

the integral from 3 to x of f(t) is

(-3/2)*x^2 + 4x + 3/2

These two values are equal, so

(-3/2)*x^2 + (4 + k)*x + 11/2 + k = (-3/2)*x^2 + 4x + 3/2

If we simplify this equation,

k*x = -k - 4

If we solve this equation to k,

k = 4/(x+1)

d/dx the integral from -1 to x of f(t)dt is

4 - 3x

Integral of 4-3x between -1 to x is ( 4x-3x^2/2) at x-( 4x-3x^2/2) at x=-1 with k added is 4(x+1)-(3/2)(x^2-(-1)^2)+k-------(1)

The integral between 3 to x is 4(x-3)-(3/2)(x^2-3^2)-----(2)

From (1) and (2) by equating,and simplifying,

4x+4-(3/2)x^2 +3/2 + k= 4x-12+(3/2)x^2+27/2. Canceling the same appearing on both sides, we get,

4+3/2+k=-12+27/2 or

**k=-4.**

Integral from -1 to x of f(x)dx is 4x-(3/2)x^2 is 4-3x^2+ K=-3

Therefore, d/dx of {4x-(3/2)x^2) +k} = 4-3x+0 =** 4-3x**.

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