# Given f(x)=(5/2)*x*(x+2)^1/2, find a,b,c if F(x)=(ax^2+bx+c)*(x+2)^1/2 is the antiderivative of f(x).

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Given f(x)=(5/2)*x*(x+2)^1/2, find a,b,c if F(x)=(ax^2+bx+c)*(x+2)^1/2 is the antiderivative of f(x).

Since F(x) = (ax^2+bx+c)(x+2)^(1/2) is the anti derivative of f(x)we get its derivative.

F'(x)= f(x) ={ (ax^2+bx+c)(x+2)^(1/2)}'

f(x) = (ax^2+bx+c) {(x+2)^(1/2)}'+ (ax^2+bx+c)'(x+2)^(1/2).

f(x) = (ax^2+bx+c)(1/2)(x+2)^(-1/2) + (2ax+b)(x+2)^(1/2)

f(x) = {(ax^2+bx+c)+2(2ax+b)(x+2)}/2(x+2)^(1/2)

((5/2)x(x+2)^1/2 = {(ax^2+bx+c)+(2ax+b)(x+2)}/(x+2)^(1/2)

5x(x+2) = 2 {(ax^2+bx+c)+(2ax+b)(x+2)}

5x^2 +10x = 2ax^2 +2bx +2c +4ax^2+8ax+2bx+4b}

5x^2+10x = 6ax^2 +(8a+4b)x+(2c+4b)

Equating the coefficients of x^2 , x and constant terms on both sides, we get:

6a = 5. So a = 4/6 = 2/3.

8a+4b = 10. So 4b = 10-8a = 10-8(2/3) = 14/3. So b = (14/3)/4 = 7/6.

0 = 2c+4b. So 2c = 0-4b = 0-4*7/6 = -28/6 = -14/3.

So c = -14/6 = -7/3..

Therefore a = 2/3, b= 7/6 and c= -7/3.

The function F(x) is the antiderivative of f(x), if and only if F'(x) = f(x).

We'll apply the product rule to determine the first derivative of F(x):

F'(x) = {(ax^2+bx+c)'*sqrt(x+2) + (ax^2+bx+c)*[sqrt(x+2)]'}

F'(x) = (2ax + b)*sqrt(x+2) + (ax^2+bx+c)/2[sqrt(x+2)]

F'(x) = (5/2)*x*sqrt(x+2)

(2ax + b)*sqrt(x+2) + (ax^2+bx+c)/2[sqrt(x+2)]=5xsqrt(x+2)/2

5(a-1)x^2 + (8a+3b - 10)x + 4b + c = 0

We'll get the system:

a - 1 = 0

a = 1 (1)

8a+3b - 10 = 0 (2)

4b + c = 0 (3)

8 + 3b - 10 = 0

3b = 2

b = 2/3

c = -8/3

**The function F(x) is:**

**F(x) = (x^2 + 2x/3 - 8/3)*sqrt(x+2)**