# What are a,b,c if f(0) = 2 , f'(1) = 26 and the definite integral of f(x) for x = 0 to x =1 is 4?f(x) = ax^4 + bx^2 +c

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### 2 Answers

f(x) = ax^4 + bx^2 + c

Given :

f(0) = 2

f'(1) 26

intg f(1) - intg f(0) = 4

We start from f(0) = 2

Let us subsitute:

f(x) = ax^4 + bx^2 + c

f(0) = 0 + 0 + c = 2

==> c= 2

==> f(x) = ax^4 + bx^2 + 2

==> f(1) = a + b + 2 = 26

==> a + b = 24

==> a= 24- b ..............(1)

Let us integrate f(x):

F(x) = intg f(x) = intg ax^4 + bx^2 + 2 dx

= (a/5)x^5 + (b/3)x^3 + 2x + C

F(0) = C

F(1) = (a/5) + (b/3) + 2 + C

Given F(1) - F(0 ) = 4

==> (a/5) + (b/3) + 2 = 4

==> (a/5) + (b/3) = 2

Multiply by 15:

==> 3a + 5b = 30................(2)

Now from (1) : a= 24-b

==> 3(24-b) + 5b = 30

==> 72 - 3b + 5b = 30

==> 2b = -42

**==> b= -21**

**==> a= 24-b = 24+21 = 45**

**==> a= 45**

### User Comments

To calculate f(0), we'll substitute x by 0 in the expression of f(x):

f(0) = a*0^4 + b*0^2 +c

f(0) = c

But f(0) = 2 (from enunciation) => **c = 2**

Now, we'll calculate f'(x):

f'(x) = (ax^4 + bx^2 +c)'

f'(x) = 4ax^3 + 2bx

But f'(1) = 26

We'll calculate f'(1) substituting x by 1 in the expression of the first derivative:

f'(1) = 4a*1^3 + 2b*1

f'(1) = 4a + 2b

But f'(1) = 26 => 4a + 2b = 26

We'll divide by 2:

2a + b = 13

b = 13 - 2a (1)

Now, we'll calculate the definite integral:

Int f(x) dx = Int (ax^4 + bx^2 +c)dx = F(1) - F(0)

Int (ax^4 + bx^2 +c)dx = aInt x^4dx + bInt x^2dx + cInt dx

Int (ax^4 + bx^2 +c)dx = a*x^5/5 + b*x^3/3 + cx

F(1) = a/5 + b/3 + c

F(0) = 0

But Int f(x)dx = 4 => a/5 + b/3 + c = 4

3a + 5b + 15c = 60

But c = 2=> 3a + 5b = 60 - 30

3a + 5b = 30 (2)

We'll substitute (1) in (2):

3a + 5(13 - 2a) = 30

3a + 65 - 10a = 30

-7a = -35

**a = 5**

b = 13 - 10

**b = 3**

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

**f(x) = 5x^4 + 3x^2 + 2**