Let a + b = 4, where a 0 for all x, prove

Q: Wait, why do we need to introduce $t$? Can't we just differentiate with respect to $a$?

You could, but $t$ makes the 'distance' concept explicit. By defining $t = b - a$, we transform a two-variable problem into a single-variable problem, which makes the derivative much cleaner and the physical interpretation of 'increasing distance' much more intuitive.

Q: Why is the derivative of the upper limit $\frac{4-t}{2}$ equal to $-\frac{1}{2}$?

It is a simple linear derivative! Since $\frac{4-t}{2} = 2 - \frac{t}{2}$, differentiating with respect to $t$ gives $\frac{d}{dt}(2) - \frac{d}{dt}(\frac{t}{2}) = 0 - \frac{1}{2} = -\frac{1}{2}$. Always remember to apply the chain rule to the upper limit!

Q: How does $g'(x) > 0$ guarantee $g(b) > g(a)$?

The condition $\frac{dg}{dx} > 0$ tells us that $g(x)$ is a strictly increasing function. Because we established that $b > 2 > a$, it follows that $b > a$. For any strictly increasing function, if the input $b$ is greater than the input $a$, the output $g(b)$ must be greater than $g(a)$.

Question

Let $a + b = 4$, where $a < 2$, and let $g(x)$ be a differentiable function. If $dg/dx > 0$ for all $x$, prove that $\int_0^a g(x) dx + \int_0^b g(x) dx$ increases as $(b - a)$ increases.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a perfectly balanced seesaw. The pivot point is at $x = 2$. We are given two points, $a$ and $b$, such that $a + b = 4$.

If you divide this equation by two, you get:

$$\frac{a+b}{2} = 2$$

This is the geometric soul of our problem: the midpoint of $a$ and $b$ is fixed at $2$. Since we know $a < 2$, the only way to keep the seesaw balanced is for $b$ to be greater than $2$. As $a$ moves to the left, $b$ must move to the right to maintain that sum of $4$.

### Defining the Distance

We want to see how the sum of the areas under the curve $g(x)$ changes as the distance between $a$ and $b$ increases. Let us define this distance as a parameter $t = b - a$.

Now, we have a system of two equations: $a + b = 4$ and $b - a = t$. By adding these, we find:

$$b = \frac{4+t}{2}$$

By subtracting them, we find:

$$a = \frac{4-t}{2}$$

Now, both our boundaries are expressed in terms of the single variable $t$. This is the key to unlocking the problem.

### The Leibniz Rule: Our Mathematical Lever

We define our integral sum as:

$$I(t) = \int_0^{\frac{4-t}{2}} g(x) dx + \int_0^{\frac{4+t}{2}} g(x) dx$$

To determine if this sum increases as $t$ increases, we must find the derivative $\frac{dI}{dt}$. We use the **Leibniz Rule** for differentiation under the integral sign, which states:

$$\frac{d}{dt} \int_0^{h(t)} g(x) dx = g(h(t)) \cdot h'(t)$$

For the first integral, the upper limit is $h(t) = \frac{4-t}{2}$. Its derivative is $h'(t) = -\frac{1}{2}$. Thus, the derivative of the first part is:

$$g\left(\frac{4-t}{2}\right) \cdot \left(-\frac{1}{2}\right) = -\frac{1}{2} g(a)$$

For the second integral, the upper limit is $h(t) = \frac{4+t}{2}$, and its derivative is $h'(t) = \frac{1}{2}$. This gives us:

$$g\left(\frac{4+t}{2}\right) \cdot \left(\frac{1}{2}\right) = \frac{1}{2} g(b)$$

### The Final Synthesis

Combining these, we get:

$$\frac{dI}{dt} = \frac{1}{2} [g(b) - g(a)]$$

We are given that $\frac{dg}{dx} > 0$, which means $g(x)$ is a **strictly increasing function**. Since $b > a$, it must be that $g(b) > g(a)$, which implies $g(b) - g(a) > 0$.

Therefore, $\frac{dI}{dt} > 0$. A positive derivative means that as $t$ increases, our integral sum $I(t)$ must also increase. You have just navigated the geometry of the seesaw and the power of the Leibniz Rule to prove a fundamental property of integrals.

Animated Solution for Mathematics - Definite Integration: Let $a + b = 4$ , where $a < 2$ , and let $g (x)$ be a differentiable function. If $d g / d x > 0$ for all $x$ , prove that $\int_{0}^{a} g (x) d x + \int_{0}^{b} g (x) d x$ increases as $(b - a)$ increases.

Visualized Solution

Understanding the Constraint $a + b = 4$

Defining the Parameter $t = b - a$

Expressing $a$ and $b$ in terms of $t$

Defining the Integral Function $I (t)$

Applying the Leibniz Rule

Differentiating the First Integral

Differentiating the Second Integral

Combining the Derivatives

Using the Monotonicity of $g (x)$

Final Conclusion

The Sigma Insight: Newton-Leibniz & Reduction Formulas

Analyzing the Setup

Defining the Distance

The Leibniz Rule

The Final Synthesis

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Visualized Solution

Understanding the Constraint .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a+b=4

Defining the Parameter .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t=b−a

Defining the Integral Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I(t)