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