Todd Bates
01/08/2024 · Middle School
Suppose that \( F(x)=\int_{1}^{x} f(t) d t \), where \[ f(t)=\int_{1}^{t^{2}} \frac{\sqrt{5+u^{5}}}{u} d u \] Find \( F^{\prime \prime}(2) \) \( F^{\prime \prime}(2)= \)
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Tutor-Verified Answer
Step-by-step Solution
To find \( F^{\prime \prime}(2) \), we start with the function \( F(x) \) defined as:
\[
F(x) = \int_{1}^{x} f(t) \, dt
\]
where
\[
f(t) = \int_{1}^{t^2} \frac{\sqrt{5 + u^5}}{u} \, du.
\]
First, we need to compute the first derivative \( F'(x) \):
\[
F'(x) = f(x).
\]
Next, we need to compute \( f'(t) \) to find \( F''(x) \):
\[
f(t) = \int_{1}^{t^2} \frac{\sqrt{5 + u^5}}{u} \, du.
\]
To differentiate \( f(t) \), we apply the Fundamental Theorem of Calculus and the chain rule. We have:
\[
f'(t) = \frac{d}{dt} \left( \int_{1}^{t^2} \frac{\sqrt{5 + u^5}}{u} \, du \right) = \frac{\sqrt{5 + (t^2)^5}}{t^2} \cdot \frac{d}{dt}(t^2) = \frac{\sqrt{5 + t^{10}}}{t^2} \cdot 2t = \frac{2\sqrt{5 + t^{10}}}{t}.
\]
Thus, we have:
\[
F''(x) = f'(x) = \frac{2\sqrt{5 + x^{10}}}{x}.
\]
Now, we need to evaluate \( F''(2) \):
\[
F''(2) = \frac{2\sqrt{5 + 2^{10}}}{2} = \sqrt{5 + 1024} = \sqrt{1029}.
\]
Therefore, the final result is:
\[
\boxed{\sqrt{1029}}.
\]
Quick Answer
\( F''(2) = \sqrt{1029} \).
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