Let I = ∫ (a to b) (x⁴ - 2x²) dx. If I is minimum then the ordered pair (a, b) is : (1) (-√2, 0) (2)

Q: Wait, why do we reject the negative root here?

Actually, we don't reject any roots! The roots are $-\sqrt{2}$, $0$, and $\sqrt{2}$. The function is negative in the interval $(-\sqrt{2}, \sqrt{2})$. We include the entire interval because every point where the function is negative contributes to making the integral smaller. If we only took a part of it, we wouldn't be capturing the full 'negative goldmine'.

Q: Why can't we just use energy conservation?

Energy conservation is a principle in physics, but here we are dealing with a pure calculus problem. The 'energy' of the integral is its value. We are minimizing the value of the integral by choosing the limits $a$ and $b$ that encompass the most negative area. It's a geometric optimization problem, not a physical one.

Q: What if the function was always positive? Would the minimum be zero?

That's a fantastic intuition! If the function were always positive, the integral would always be positive. To minimize it, you would want the interval to be as small as possible, ideally a single point where the integral becomes zero. But here, our function dips below the axis, so we have the luxury of making the integral negative!

Question

Let $I = \int_a^b (x^4 - 2x^2) dx$. If I is minimum then the ordered pair (a, b) is :

Options:

(A) (-\sqrt{2}, 0)

(B) (-\sqrt{2}, \sqrt{2})

(C) (0, \sqrt{2})

(D) (\sqrt{2}, -\sqrt{2})

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler of the mathematical landscape! Today, we are going to unravel a problem that might look like a standard integration exercise, but it is actually a beautiful lesson in the visual nature of calculus.

We are tasked with minimizing the integral:

$$I = \int_a^b (x^4 - 2x^2) dx$$

At first glance, you might be tempted to jump straight into finding the antiderivative, but let's pause. Calculus is not just about crunching numbers; it is about understanding the geometry of the functions we manipulate.

### The Signed Area Philosophy

Imagine the function $f(x) = x^4 - 2x^2$ as a roller coaster track. The definite integral $I = \int_a^b f(x) dx$ is not just a number; it is the signed area between this track and the $x$-axis.

When the track is above the $x$-axis, the area is positive, adding to our total. When the track dips below the $x$-axis, the area is negative, subtracting from our total.

Our goal is to minimize $I$. To make a sum as small as possible, we want to accumulate as much negative value as we can while avoiding any positive value. It is like trying to fill a bucket with negative numbers—we want to scoop up every bit of the negative region and leave the positive region untouched.

### Unmasking the Curve

To find where our "negative goldmine" lies, we need to know where the curve crosses the $x$-axis. We set $f(x) = 0$:

$$x^4 - 2x^2 = 0$$

Factoring this is straightforward: $x^2(x^2 - 2) = 0$. This gives us the roots $x = 0$ (a double root) and $x = \pm\sqrt{2}$.

These roots divide the $x$-axis into three distinct regions: $x < -\sqrt{2}$, $-\sqrt{2} < x < \sqrt{2}$, and $x > \sqrt{2}$.

Now, let's test the sign of $f(x)$ in these regions. In the middle interval $(-\sqrt{2}, \sqrt{2})$, if we pick a value like $x = 1$, we get:

$$f(1) = 1^4 - 2(1)^2 = 1 - 2 = -1$$

The function is negative! This is exactly what we want. In the outer regions, $x^2$ is always positive, and $x^2 - 2$ is also positive, making $f(x)$ positive.

If we were to include these regions in our integral, we would be adding positive area, which would increase the value of $I$. We must avoid them at all costs.

### The Final Strategy

To minimize $I$, we must integrate exactly over the interval where $f(x) \le 0$. This interval is $(-\sqrt{2}, \sqrt{2})$.

By choosing $a = -\sqrt{2}$ and $b = \sqrt{2}$, we capture the entire negative region and exclude all positive regions. Any other choice of $a$ and $b$ would either miss some of the negative area or include some of the positive area, both of which would result in a larger (less negative) value for $I$.

Therefore, the ordered pair $(a, b)$ that minimizes the integral is **$(-\sqrt{2}, \sqrt{2})$**.

You have successfully navigated the geometry of the function and found the optimal limits. Keep this visual approach in your toolkit—it will serve you well in the most challenging JEE problems!

Animated Solution for Mathematics - Definite Integration: Let $I = \int_{a}^{b} (x^{4} - 2 x^{2}) d x$ . If I is minimum then the ordered pair (a, b) is :

Select Answer:

Visualized Solution

The Objective

Geometric Meaning of Definite Integral

Strategy to Minimize $I$

Finding the Roots

Calculating the Roots

Sign Analysis: The Negative Region

Sign Analysis: The Positive Regions

Concluding the Limits

Verifying the Options

The Sigma Insight: Area Bounded by Curves

Analyzing the Setup

The Signed Area Philosophy

Unmasking the Curve

The Final Strategy

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Animated Solution for Mathematics - Definite Integration: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let I=∫ab​(x4−2x2)dx. If I is minimum then the ordered pair (a, b) is :