Let's break down the problem step by step.

We are tasked with evaluating the following integral: $I = \int_0^{\pi/2} \cos(x) f'''(\sin(x)) \, dx$ where $f$ is a three-times differentiable function with its values and the values of its derivatives given in a table.

Step 1: Substitution

We will use the substitution $u = \sin(x)$. Thus: $du = \cos(x) \, dx$ When $x = 0$, $u = \sin(0) = 0$, and when $x = \pi/2$, $u = \sin(\pi/2) = 1$.

Thus, the integral transforms into: $I = \int_0^1 f'''(u) \, du$

Step 2: Use of Given Table

From the table, we have the following information about $f'''(u)$:

• $f'''(0) = 3$
• $f'''(1) = -4$

Since we are evaluating an integral over $f'''(u)$ from 0 to 1, we can approximate this using a linear interpolation (trapezoidal rule), since the exact form of $f'''(u)$ is unknown. The integral becomes: [ I \approx \frac{1}{2} \left[ f'''(0) + f'''(1) \right] = \frac{1}{2} \left[ 3 + (-4) \right] = \frac{1}{2} \cdot (-1) = -\frac{1}{2} ]

Thus, the value of the integral is approximately $-\frac{1}{2}$.

Final Answer:

$I = -\frac{1}{2}$

Would you like further details or have any questions?

Here are 5 related questions for practice:

1. How would the result change if the table values for $f'''(x)$ were different?
2. How does the substitution $u = \sin(x)$ simplify the integral in this context?
3. What other methods could be used to evaluate this integral?
4. Why is it acceptable to use linear interpolation in this case?
5. What would the result be if $f'''(u)$ were a constant function?

Tip: When solving integrals with unknown functions, tables of values can often be handled using approximation techniques like the trapezoidal or Simpson’s rule.