Animated Solution for Mathematics - Definite Integration: If ${\int }_{-0.15}^{0.15}|100x^2-1|dx=\frac{k}{3000}$, then $k$ is equal to \_\_\_\_\_.

Welcome, fellow traveler on the path to JEE mastery. Today, we are going to dissect a problem that at first glance might seem like a standard calculus exercise, but beneath the surface, it is a beautiful dance of symmetry and piecewise definition.

We are tasked with evaluating the integral:

The modulus function, $|100x^2-1|$, acts like a mirror. Imagine the standard parabola $y=100x^2-1$. It dips below the x-axis, creating a valley. The modulus operator takes that negative dip and reflects it upward, creating a distinct 'W' shape. This 'W' is our playground for today.

Before we dive into the heavy algebra, let us pause and observe the limits of integration: from $-0.15$ to $0.15$. Whenever you see symmetric limits, your first instinct should be to check for symmetry in the function.

Is it even? Is it odd? Let us test $f(x)=|100x^2-1|$. Replacing $x$ with $-x$, we get:

It is an even function! This is a gift. It means the area on the left side of the y-axis is a perfect reflection of the area on the right. We can simplify our life by calculating the integral from $0$ to $0.15$ and simply doubling the result:

Now, we must confront the modulus. We cannot integrate $|100x^2-1|$ directly. We need to know where the expression inside changes sign.

We set $100x^2-1=0$, which leads us to $x^2=\frac{1}{100}$, or $x=\pm 0.1$. Since our integral is now restricted to the interval $[0,0.15]$, our critical point of interest is $x=0.1$.

For $x$ in the range $[0,0.1]$, the parabola $100x^2-1$ is negative, so the modulus flips it to $1-100x^2$. For $x$ in the range $[0.1,0.15]$, the parabola is already positive, so the modulus leaves it as $100x^2-1$.

With our critical point identified, we split the integral:

Now, we apply the fundamental power rule of integration: $\int x^{n}dx=\frac{x^{n+1}}{n+1}$. The first part becomes:

The second part becomes:

Evaluating these requires precision. For the first part, substituting $0.1$ gives:

For the second part, we calculate the values at $0.15$ and $0.1$:

Simplifying this yields $\frac{0.0875}{3}$.

We are at the finish line. Adding our two results together, we get:

The problem states that $I=\frac{k}{3000}$. Equating our result, $\frac{0.575}{3}=\frac{k}{3000}$, we multiply both sides by $1000$ to align the denominators:

Thus, $k=575$. You have navigated the symmetry, the modulus, and the integration with precision. Take a moment to appreciate the elegance of the result—a testament to your hard work and conceptual clarity.