Let g(x) = ∫₀ˣ f(t)dt, where f is continuous function in [0, 3] such that 1/3 ≤ f(t) ≤ 1 for all t ∈

Q: Wait, why do we split the integral at $t=1$?

Great question! We split the integral because the definition of the function $f(t)$ changes at $t=1$. In calculus, when a function's behavior is defined piecewise, the most natural way to analyze its integral is to break it down into those specific pieces. It allows us to apply the given bounds accurately to each region.

Q: Why can't we just take the average value of the function to find the integral?

That is a common misconception! The average value theorem gives us the value of the integral if we know the function is continuous, but here we don't know the exact function—only its bounds. We are looking for the *range* of possible values for the integral, which requires us to find the absolute minimum and maximum possible areas under the curve, not just an average.

Q: Is it possible for $g(3)$ to be exactly $1/3$ or $2$?

Yes, absolutely! If $f(t)$ is a constant function equal to its lower bound in both intervals, the integral will equal the lower bound of the total range. Similarly, if $f(t)$ is a constant function equal to its upper bound in both intervals, the integral will equal the upper bound. The interval is closed because these constant functions are valid candidates for $f(t)$.

Question

Let $g(x) = \int_0^x f(t) dt$, where $f$ is continuous function in $[0, 3]$ such that $\frac{1}{3} \le f(t) \le 1$ for all $t \in [0, 1]$ and $0 \le f(t) \le \frac{1}{2}$ for all $t \in (1, 3]$. The largest possible interval in which $g(3)$ lies is:

Options:

(A) $[-1, -1/2]$

(B) $[-3/2, -1]$

(C) $[1/3, 2]$

(D) $[1, 3]$

Sigma Percentile · Accepted Answer

### The Geometry of Uncertainty: A Journey into Definite Integrals

Imagine you are standing before a vast, unknown landscape. You are tasked with finding the total area of a region, but there is a catch: you don't have a map.

You only have a set of rules describing the height of the terrain at different points. This is exactly the challenge we face with the function $g(x) = \int_0^x f(t) dt$.

We are asked to find the range of $g(3)$, which is the total area under the curve $f(t)$ from $t=0$ to $t=3$. Since $f(t)$ is not explicitly given, we must rely on the constraints provided to bound our answer.

This is not just an algebraic exercise; it is a lesson in how to navigate uncertainty using the power of calculus.

### Phase 1: The Strategy of Division

When a problem seems too large to handle, the best strategy is to divide and conquer. The function $f(t)$ changes its behavior at $t=1$.

For the first part of our journey, $t \in [0, 1]$, the function is constrained by $\frac{1}{3} \le f(t) \le 1$. For the second part, $t \in (1, 3]$, the constraints shift to $0 \le f(t) \le \frac{1}{2}$.

Because the rules change, we must split our integral using the additive property of definite integrals:

$$g(3) = \int_0^1 f(t) dt + \int_1^3 f(t) dt$$

By separating the problem, we transform one intimidating integral into two manageable pieces. We are essentially calculating the area of two separate, bounded corridors.

### Phase 2: Analyzing the First Corridor

Let's focus on the first interval, $t \in [0, 1]$. We know that $\frac{1}{3} \le f(t) \le 1$.

To find the bounds of the integral, we integrate this inequality across the interval $[0, 1]$:

$$\int_0^1 \frac{1}{3} dt \le \int_0^1 f(t) dt \le \int_0^1 1 dt$$

Calculating these simple integrals, we get $\frac{1}{3} [t]_0^1 \le \int_0^1 f(t) dt \le 1 [t]_0^1$, which simplifies to $\frac{1}{3} \le \int_0^1 f(t) dt \le 1$.

This tells us that the area under the curve in the first region is at least $\frac{1}{3}$ and at most $1$. We have successfully constrained the first part of our total area.

### Phase 3: Analyzing the Second Corridor

Now, we turn our attention to the second interval, $t \in (1, 3]$. Here, the constraints are $0 \le f(t) \le \frac{1}{2}$.

Again, we integrate this inequality, but be careful with the limits—the width of this interval is $3 - 1 = 2$:

$$\int_1^3 0 dt \le \int_1^3 f(t) dt \le \int_1^3 \frac{1}{2} dt$$

This yields $0 \le \int_1^3 f(t) dt \le \frac{1}{2} [t]_1^3$, which simplifies to $0 \le \int_1^3 f(t) dt \le \frac{1}{2} (3 - 1) = 1$.

The area in the second region is at least $0$ and at most $1$. Notice how the width of the interval significantly impacts the potential area!

### Phase 4: The Synthesis

We have our two pieces. To find the total range of $g(3)$, we simply add the minimums and the maximums together.

The minimum possible value for $g(3)$ is the sum of the minimums: $\frac{1}{3} + 0 = \frac{1}{3}$. The maximum possible value is the sum of the maximums: $1 + 1 = 2$.

Therefore, the total integral $g(3)$ must lie in the interval **$[\frac{1}{3}, 2]$**.

This result is elegant and powerful. We didn't need to know the exact path of $f(t)$; we only needed to know its boundaries.

By breaking the problem down and respecting the constraints of each region, we found the exact range of possibilities. Keep this mindset as you tackle more complex problems—often, the secret lies in how you choose to divide the challenge.

Select Answer:

Visualized Solution

Understanding $g (3)$

Splitting the Integral

Bounds for $t \in [0, 1]$

Integrating First Interval

Bounds for $t \in (1, 3]$

Integrating Second Interval

Combining the Results

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

The Geometry of Uncertainty

Phase 1

Phase 2

Phase 3

Phase 4

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