The correct answer is 3π²/8.

Step-by-step Explanation:

From the given image, we need to evaluate the following limit:
$L=\underset{x\to {\frac{\pi }{2}}^{-}}{lim}\frac{{\int }_{x^3}^{{\left(\frac{\pi }{2}\right)}^3}cos\left(t^{1/3}\right)dt}{{\left(x-\frac{\pi }{2}\right)}^2}$

As $x\to {\frac{\pi }{2}}^{-}$, the lower limit of integration $x^3$ approaches ${\left(\frac{\pi }{2}\right)}^3$.
Therefore, the numerator approaches:
${\int }_{{\left(\frac{\pi }{2}\right)}^3}^{{\left(\frac{\pi }{2}\right)}^3}cos\left(t^{1/3}\right)dt=0$

The denominator also approaches:
${\left(\frac{\pi }{2}-\frac{\pi }{2}\right)}^2=0$

Since this is an indeterminate form of type $\frac{0}{0}$, we can apply L'Hôpital's Rule by differentiating the numerator and the denominator with respect to $x$.

1. Differentiating the Numerator:
Using the Leibniz Rule for differentiation under the integral sign:
$\frac{d}{dx}\left[{\int }_{x^3}^{{\left(\frac{\pi }{2}\right)}^3}cos\left(t^{1/3}\right)dt\right]=-cos\left({\left(x^3\right)}^{1/3}\right)·\frac{d}{dx}\left(x^3\right)$
$=-3x^{2}cos\left(x\right)$

2. Differentiating the Denominator:
$\frac{d}{dx}\left[{\left(x-\frac{\pi }{2}\right)}^2\right]=2\left(x-\frac{\pi }{2}\right)$

3. Applying the Limit:
Substitute the derivatives back into the limit expression:
$L=\underset{x\to {\frac{\pi }{2}}^{-}}{lim}\frac{-3x^{2}cos\left(x\right)}{2\left(x-\frac{\pi }{2}\right)}$

We can separate the factors that do not evaluate to 0:
$L=\left[\underset{x\to {\frac{\pi }{2}}^{-}}{lim}\frac{-3x^2}{2}\right]·\left[\underset{x\to {\frac{\pi }{2}}^{-}}{lim}\frac{cos\left(x\right)}{x-\frac{\pi }{2}}\right]$

Evaluate the first part:
$\underset{x\to {\frac{\pi }{2}}^{-}}{lim}\frac{-3x^2}{2}=\frac{-3{\left(\frac{\pi }{2}\right)}^2}{2}=-\frac{3{\pi }^2}{8}$

Evaluate the second part using L'Hôpital's Rule again because it is a $\frac{0}{0}$ form:
$\underset{x\to {\frac{\pi }{2}}^{-}}{lim}\frac{cos\left(x\right)}{x-\frac{\pi }{2}}$

Combine the results:
$L=\left(-\frac{3{\pi }^2}{8}\right)·(-1)=\frac{3{\pi }^2}{8}$