Let f(x) be a continuous function such that f(1) = 3 and f '(x) = sqrt( x^3 + 4 ). what is the value of f(5)?

The value of f(5) is 49.1

Step-by-step explanation:

To find f(x) from f'(x) use the integration

f(x) = ∫ f'(x)

1. Find The integration of f'(x) with the constant term

2. Substitute x by 1 and f(x) by π to find the constant term

3. Write the differential function f(x) and substitute x by 5 to find f(5)

∵ f'(x) =  + 6

- Change the root to fraction power

∵  =

∴ f'(x) =  + 6

∴ f(x) = ∫  + 6

- In integration add the power by 1 and divide the coefficient by the

 new power and insert x with the constant term

∴ f(x) =  + 6x + c

- c is the constant of integration

∵

∴ f(x) =   + 6x + c

- To find c substitute x by 1 and f(x) by π

∴ π =   + 6(1) + c

∴ π =  + 6 + c

∴ π = 6.4 + c

- Subtract 6.4 from both sides

∴ c = - 3.2584

∴ f(x) =   + 6x - 3.2584

To find f(5) Substitute x by 5

∵ x = 5

∴ f(5) =   + 6(5) - 3.2584

∴ f(5) = 49.1

49%

step-by-step explanation:

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