Given a function f(x) such that (i) it is integrable over every interval on the real line and (ii) f

Q: Why do we have to differentiate the integral? Can't we just say it's obvious from the graph?

In mathematics, intuition is our compass, but proof is our map. While the graph makes it look obvious, the derivative provides the rigorous, undeniable proof that the rate of change is zero. In JEE Advanced, we don't just want to see the answer; we want to prove the mechanism behind it.

Q: Is the Leibniz Rule the only way to solve this? It feels like a heavy hammer for a small nail.

It is indeed a powerful tool, but it is the most elegant one here. You could use substitution $u = x - a$, which would also work, but Leibniz cuts straight to the heart of the matter by directly relating the change in the integral to the change in the boundaries. It is a skill that will serve you well in much harder problems.

Q: What if the function wasn't periodic? Would the integral still be independent of $a$?

That is a fantastic question! If the function were not periodic, the integral would almost certainly change as you slide the window. The independence of $a$ is a unique property born directly from the periodicity condition $f(t+x) = f(x)$. Without that symmetry, the area would fluctuate wildly.

Question

Given a function $f(x)$ such that (i) it is integrable over every interval on the real line and (ii) $f(t+x) = f(x)$, for every $x$ and a real $t$, then show that the integral $\int_a^{a+t} f(x) dx$ is independent of $a$.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler of the mathematical landscape. Today, we are not just solving a problem; we are uncovering a fundamental truth about the nature of periodic functions.

Imagine you are standing on a vast, infinite plain. In front of you, a wave-like function $f(x)$ stretches to the horizon. It is a beautiful, rhythmic pattern, repeating itself exactly every $t$ units, defined by the property:
$$f(x+t) = f(x)$$

Now, consider a window of width $t$. You place this window on the graph, starting at some point $a$, and ending at $a+t$. The area under the curve within this window is our integral:
$$I(a) = \int_a^{a+t} f(x) dx$$

The question we face is profound: Does the area change if we slide this window? Does the 'amount' of function we capture depend on where we start?

### Phase 1: The Sliding Window

When we define $I(a) = \int_a^{a+t} f(x) dx$, we are essentially creating a 'sliding window' of fixed width $t$. Geometrically, this represents the area under the curve $f(x)$ bounded between the vertical lines $x = a$ and $x = a+t$.

As we vary $a$, we are sliding this window along the $x$-axis. Our intuition tells us that because the function repeats itself, the area should remain constant.

In the world of JEE Advanced, intuition must be backed by the ironclad logic of calculus. We need to prove that the rate of change of this area is zero. If the rate of change is zero, the area is constant.

### Phase 2: The Calculus Weapon

To differentiate an integral where the limits themselves depend on the variable $a$, we need a very powerful tool from our mathematical arsenal: the **Leibniz Integral Rule**. This rule is the scalpel of calculus.

The rule states:
$$\frac{d}{da} \int_{g(a)}^{h(a)} f(x) dx = f(h(a)) \cdot h'(a) - f(g(a)) \cdot g'(a)$$

Here, our lower limit is $g(a) = a$ and our upper limit is $h(a) = a+t$. This rule is not just a formula; it is a statement about how the area 'grows' at the leading edge and 'shrinks' at the trailing edge as we slide the window.

### Phase 3: The Clash of Titans

Now, let us apply this rule to our integral $I(a)$. We calculate the derivative:
$$I'(a) = \frac{d}{da} \int_a^{a+t} f(x) dx$$

Applying the Leibniz Rule, we get:
$$I'(a) = f(a+t) \cdot \frac{d}{da}(a+t) - f(a) \cdot \frac{d}{da}(a)$$

Since $t$ is a constant, the derivative of $a+t$ with respect to $a$ is simply $1$, and the derivative of $a$ with respect to $a$ is also $1$. This simplifies our expression beautifully to:
$$I'(a) = f(a+t) - f(a)$$

### The Final Revelation

We are given the condition of periodicity: $f(t+x) = f(x)$ for all $x$. If we substitute $x = a$ into this condition, we get $f(a+t) = f(a)$.

Look at what happens to our derivative expression:
$$I'(a) = f(a) - f(a) = 0$$

Because the derivative of $I(a)$ is zero for all values of $a$, the function $I(a)$ must be a constant. It does not matter where you place your window of width $t$; the area captured will always be the same.

We have proven that the integral $\int_a^{a+t} f(x) dx$ is independent of $a$. This is the beauty of mathematics—taking a complex, dynamic system and finding the underlying stillness within it.

Keep this geometric insight with you: the area under a periodic function over exactly one period is an **invariant**, a constant truth in a changing world.

Animated Solution for Mathematics - Definite Integration: Given a function $f (x)$ such that (i) it is integrable over every interval on the real line and (ii) $f (t + x) = f (x)$ , for every $x$ and a real $t$ , then show that the integral $\int_{a}^{a + t} f (x) d x$ is independent of $a$ .

Visualized Solution

Visualizing the Periodic Function $f (x)$

Defining the Integral $I (a)$

Formulating the Goal

Introducing the Leibniz Integral Rule

Applying Leibniz Rule to $I (a)$

Differentiating the Limits

Utilizing Periodicity

Final Conclusion

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Analyzing the Setup

Phase 1

Phase 2

Phase 3

The Final Revelation

Similar Questions

If $f (a + b + 1 - x) = f (x) \forall x$ , where $a$ and $b$ are fixed positive real numbers, then $\frac{1}{( a + b )} \int_{a}^{b} x (f (x) + f (x + 1)) d x$ is equal to

If $f$ and $g$ are continuous function on $[0, a]$ satisfying $f (x) = f (a - x)$ and $g (x) + g (a - x) = 2$ , then show that $\int_{0}^{a} f (x) g (x) d x = \int_{0}^{a} f (x) d x$

If $f (a + b + 1 - x) = f (x)$ , for all $x$ , where $a$ and $b$ are fixed positive real numbers, then $\frac{1}{( a + b )} \int_{a}^{b} x (f (x) + f (x + 1)) d x$ is equal to

If $f (a + b - x) = f (x)$ then $\int_{a}^{b} x f (x) d x$ is equal to

Let $f (x) = \int_{0}^{x} g (t) d t$ , where g is a non-zero even function. If $f (x + 5) = g (x)$ , then $\int_{0}^{x} f (t) d t$ equals-

Let $T > 0$ be a fixed real number. Suppose $f$ is a continuous function such that for all $x \in R$ , $f (x + T) = f (x)$ . If $I = \int_{0}^{T} f (x) d x$ then the value of $\int_{3}^{3 + 3 T} f (2 x) d x$ is

The value of $\int_{1}^{a} [x] f^{'} (x) d x, a > 1$ where $[x]$ denotes the greatest integer not exceeding $x$ is

For $x > 0$ , let $f (x) = \int_{e}^{x} \frac{l n t}{1 + t} d t$ . Find the function $f (x) + f (\frac{1}{x})$ and show that $f (e) + f (\frac{1}{e}) = \frac{1}{2}$ .

Let $f : R \to R$ be continuous function satisfying $f (x) + f (x + k) = n$ , for all $x \in R$ where $k > 0$ and $n$ is a positive integer. If $I_{1} = \int_{0}^{4 nk} f (x) d x$ and $I_{2} = \int_{- k}^{3 k} f (x) d x$ , then

Show that $\int_{0}^{π} x f (sin x) d x = \frac{π}{2} \int_{0}^{π} f (sin x) d x$ .

Visualized Solution

Visualizing the Periodic Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

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

Formulating the Goal

Introducing the Leibniz Integral Rule

Applying Leibniz Rule to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I(a)

Differentiating the Limits

Utilizing Periodicity

Final Conclusion

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Analyzing the Setup

Phase 1

Phase 2

Phase 3

The Final Revelation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f(a+b+1−x)=f(x)∀x, where a and b are fixed positive real numbers, then (a+b)1​∫ab​x(f(x)+f(x+1))dx is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f and g are continuous function on [0,a] satisfying f(x)=f(a−x) and g(x)+g(a−x)=2, then show that ∫0a​f(x)g(x)dx=∫0a​f(x)dx

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f(a+b+1−x)=f(x), for all x, where a and b are fixed positive real numbers, then (a+b)1​∫ab​x(f(x)+f(x+1))dx is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f(a+b−x)=f(x) then ∫ab​xf(x)dx is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x)=∫0x​g(t)dt, where g is a non-zero even function. If f(x+5)=g(x), then ∫0x​f(t)dt equals-

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let T>0 be a fixed real number. Suppose f is a continuous function such that for all x∈R, f(x+T)=f(x). If I=∫0T​f(x)dx then the value of ∫33+3T​f(2x)dx is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of ∫1a​[x]f′(x)dx,a>1 where [x] denotes the greatest integer not exceeding x is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For x>0, let f(x)=∫ex​1+tlnt​dt. Find the function f(x)+f(x1​) and show that f(e)+f(e1​)=21​.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f:R→R be continuous function satisfying f(x)+f(x+k)=n, for all x∈R where k>0 and n is a positive integer. If I1​=∫04nk​f(x)dx and I2​=∫−k3k​f(x)dx, then

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Show that ∫0π​xf(sinx)dx=2π​∫0π​f(sinx)dx.