Let $f(x) = \int_0^x g(t) dt$, where g is a non-zero even function. If $f(x+5) = g(x)$, then $\int_0^x f(t) dt$ equals- Options: (A) $\int_{x+5}^5 g(t) dt$ (B) $\int_{x+5}^5 5 g(t) dt$ (C) $\int_5^{x+5} g(t) dt$ (D) $\int_5^{x+5} 2 g(t) dt$

### Analyzing the Symmetry of the Function We begin with a function $g(x)$ that is symmetric across the $y$-axis, meaning it is an even function such that $g(-x) = g(x)$. We define $f(x)$ as the accumulation function: $$f(x) = \int_0^x g(t) dt$$ To determine the parity of $f(x)$, we evaluate $f(-x)$: $$f(-x) = \int_0^{-x} g(t) dt$$ By substituting $t = -u$, we find $dt = -du$. As $t$ ranges from $0$ to $-x$, $u$ ranges from $0$ to $x$. Thus: $$f(-x) = \int_0^x g(-u) (-du) = -\int_0^x g(u) du = -f(x)$$ Since $f(-x) = -f(x)$, we have proven that **$f(x)$ is an odd function**. ### The Bridge: Connecting $f$ and $g$ We are given the condition $f(x+5) = g(x)$. Because $g$ is even, we know $g(x) = g(-x)$, which implies: $$f(x+5) = g(-x)$$ Replacing $x$ with $-x$ in the original equation yields $f(-x+5) = g(-x)$, which simplifies to $f(5-x) = g(x)$. Utilizing the odd property of $f$, we rewrite $f(5-x)$ as $f(-(x-5)) = -f(x-5)$. Therefore, we establish the crucial relationship: $$-f(x-5) = g(x) \Rightarrow f(x-5) = -g(x)$$ ### The Integral Transformation: The Final Act Our goal is to evaluate the integral $I = \int_0^x f(t) dt$. To utilize our established relationship, we perform the substitution $t = u - 5$, which implies $dt = du$. When $t=0$, $u=5$, and when $t=x$, $u=x+5$. The integral transforms as follows: $$I = \int_5^{x+5} f(u-5) du$$ Substituting $f(u-5) = -g(u)$ into the expression, we obtain: $$I = \int_5^{x+5} (-g(u)) du = -\int_5^{x+5} g(u) du$$ By applying the property of definite integrals that allows us to swap the limits by negating the integral, we arrive at the final result: $$I = \int_{x+5}^5 g(u) du$$ Changing the dummy variable back to $t$, the final expression is: $$\mathbf{\int_{x+5}^5 g(t) dt}$$

JEE Main 2019 (08 April Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Definite Integration: 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-

Q: Wait, why is $f(x)$ odd? I thought the integral of an even function is always even?

That is a classic trap! Remember, $f(x) = \int_0^x g(t) dt$. If $g(t)$ is even, the area from $0$ to $x$ is the negative of the area from $0$ to $-x$. Think about the geometry: if you move to the right, you add area; if you move to the left, you are essentially subtracting the same area. Thus, $f(-x) = -f(x)$, making it an odd function. It is a beautiful property of symmetry!

Q: Can we just differentiate both sides of $f(x+5) = g(x)$?

Yes, you absolutely can! Differentiating gives $f'(x+5) = g'(x)$. Since $f'(x) = g(x)$, this becomes $g(x+5) = g'(x)$. This is a valid path, but it leads to a differential equation. While elegant, the substitution method we used is often faster and less prone to calculation errors in an exam setting. Always look for the path of least resistance!

Select Answer:

Progressive Hints

Hint 1

Before diving into the algebra, pause and look at the symmetry. You are given that

g (x)

is an even function. If

f (x) = \int_{0}^{x} g (t) d t

, what happens to the area under the curve when you flip the sign of

x

? Does the area accumulate in the same direction or the opposite? This is the key to unlocking the parity of

f (x)

Hint 2

You have the relation

f (x + 5) = g (x)

. Since

g (x)

is even, we know

g (x) = g (- x)

. Use this to transform the given equation into a form involving

f (5 - x)

. Then, use the parity of

f (x)

you discovered in the first step to relate

f (5 - x)

f (x - 5)

Hint 3

Now, look at the target integral

\int_{0}^{x} f (t) d t

. To utilize the relationship

f (x - 5) = - g (x)

, perform a substitution

t = u - 5

. Don't forget to transform your limits of integration from

[0, x]

[5, x + 5]

. The rest is just a matter of applying the properties of definite integrals.

Visualized Solution

The Given Functions

Given: $f (x) = \int_{0}^{x} g (t) d t$
$g (x)$ is a non-zero even function.
This means $g (- x) = g (x)$ .

Parity of $f (x)$

Let's find the parity of $f (x)$ by evaluating $f (- x)$ .
$f (- x) = \int_{0}^{- x} g (t) d t$

Substituting to find $f (- x)$

Let $t = - u ⟹ d t = - d u$ .
Limits: when $t = 0, u = 0$ ; when $t = - x, u = x$ .
$f (- x) = \int_{0}^{x} g (- u) (- d u)$

$f (x)$ is an Odd Function

Since $g$ is even, $g (- u) = g (u)$ .
$f (- x) = - \int_{0}^{x} g (u) d u = - f (x)$
Conclusion: $f (x)$ is an odd function.

Analyzing $f (x + 5) = g (x)$

We are given: $f (x + 5) = g (x)$
Since $g (x)$ is even, we know $g (- x) = g (x)$ .
Let's substitute $- x$ into our given equation.

Substituting $- x$

Substitute $- x$ for $x$ :
$f (- x + 5) = g (- x)$
$f (5 - x) = g (x)$

Applying the Odd Property

We have $f (5 - x) = g (x)$ .
Rewrite $5 - x$ as $- (x - 5)$ .
$f (- (x - 5)) = g (x)$

Final Relation between $f$ and $g$

Using $f (- z) = - f (z)$ :
$- f (x - 5) = g (x)$
Therefore: $f (x - 5) = - g (x)$

Setting up the Target Integral

We need to evaluate: $I = \int_{0}^{x} f (t) d t$
To use our relation $f (u - 5) = - g (u)$ , we need a substitution.

Integral Substitution

Let $t = u - 5 ⟹ d t = d u$ .
Let's find the new limits for $u$ .

Transforming the Limits

Lower limit: When $t = 0 ⟹ u = 5$ .
Upper limit: When $t = x ⟹ u = x + 5$ .
The integral becomes: $I = \int_{5}^{x + 5} f (u - 5) d u$

Substituting $- g (u)$

We know $f (u - 5) = - g (u)$ .
Substitute this into the integral:
$I = \int_{5}^{x + 5} (- g (u)) d u$

Reversing Limits and Final Answer

Use the property: $\int_{a}^{b} h (x) d x = - \int_{b}^{a} h (x) d x$
$I = \int_{x + 5}^{5} g (u) d u$
Changing the dummy variable back to $t$ :
$I = \int_{x + 5}^{5} g (t) d t$

00:00 / 00:00

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Solution Diagram

Analyzing the Symmetry of the Function

We begin with a function

g (x)

that is symmetric across the

y

-axis, meaning it is an even function such that

g (- x) = g (x)

. We define

f (x)

as the accumulation function:

f (x) = \int_{0}^{x} g (t) d t

To determine the parity of

f (x)

, we evaluate

f (- x)

f (- x) = \int_{0}^{- x} g (t) d t

By substituting

t = - u

, we find

d t = - d u

. As

t

ranges from

0

- x

u

ranges from

0

x

. Thus:

f (- x) = \int_{0}^{x} g (- u) (- d u) = - \int_{0}^{x} g (u) d u = - f (x)

Since

f (- x) = - f (x)

, we have proven that $f (x)$ is an odd function.

The Bridge

Connecting

f

and

g

We are given the condition

f (x + 5) = g (x)

. Because

g

is even, we know

g (x) = g (- x)

, which implies:

f (x + 5) = g (- x)

Replacing

x

with

- x

in the original equation yields

f (- x + 5) = g (- x)

, which simplifies to

f (5 - x) = g (x)

. Utilizing the odd property of

f

, we rewrite

f (5 - x)

f (- (x - 5)) = - f (x - 5)

Therefore, we establish the crucial relationship:

- f (x - 5) = g (x) \Rightarrow f (x - 5) = - g (x)

The Integral Transformation

The Final Act

Our goal is to evaluate the integral

I = \int_{0}^{x} f (t) d t

. To utilize our established relationship, we perform the substitution

t = u - 5

, which implies

d t = d u

When

t = 0

u = 5

, and when

t = x

u = x + 5

. The integral transforms as follows:

I = \int_{5}^{x + 5} f (u - 5) d u

Substituting

f (u - 5) = - g (u)

into the expression, we obtain:

I = \int_{5}^{x + 5} (- g (u)) d u = - \int_{5}^{x + 5} g (u) d u

By applying the property of definite integrals that allows us to swap the limits by negating the integral, we arrive at the final result:

I = \int_{x + 5}^{5} g (u) d u

Changing the dummy variable back to

t

, the final expression is:

\int_{x + 5}^{5} g (t) dt

Common Traps & FAQs

Wait, why is

f (x)

odd? I thought the integral of an even function is always even?

That is a classic trap! Remember,

f (x) = \int_{0}^{x} g (t) d t

. If

g (t)

is even, the area from

0

x

is the negative of the area from

0

- x

. Think about the geometry: if you move to the right, you add area; if you move to the left, you are essentially subtracting the same area. Thus,

f (- x) = - f (x)

, making it an odd function. It is a beautiful property of symmetry!

Why did we substitute

t = u - 5

? How do we know that's the right move?

Great question! In JEE problems, we often look for a 'bridge' between the given information and the target. We had a relation involving

f (x - 5)

and

g (x)

. Our integral was in terms of

f (t)

. By setting

t = u - 5

, we transformed the integrand

f (t)

into

f (u - 5)

, which we already knew was equal to

- g (u)

. It is all about aligning your tools with the target!

Can we just differentiate both sides of

f (x + 5) = g (x)

Yes, you absolutely can! Differentiating gives

f^{'} (x + 5) = g^{'} (x)

. Since

f^{'} (x) = g (x)

, this becomes

g (x + 5) = g^{'} (x)

. This is a valid path, but it leads to a differential equation. While elegant, the substitution method we used is often faster and less prone to calculation errors in an exam setting. Always look for the path of least resistance!

Animated Solution for Mathematics - Definite Integration: 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-

Select Answer:

Visualized Solution

The Given Functions

Parity of $f (x)$

Substituting to find $f (- x)$

$f (x)$ is an Odd Function

Analyzing $f (x + 5) = g (x)$

Substituting $- x$

Applying the Odd Property

Final Relation between $f$ and $g$

Setting up the Target Integral

Integral Substitution

Transforming the Limits

Substituting $- g (u)$

Reversing Limits and Final Answer

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Analyzing the Symmetry of the Function

The Bridge

The Integral Transformation

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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 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-

Select Answer:

Visualized Solution

The Given Functions

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

Substituting to find .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(−x)

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

Analyzing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x+5)=g(x)

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } −x

Applying the Odd Property

Final Relation between .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g

Setting up the Target Integral

Integral Substitution

Transforming the Limits

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } −g(u)

Reversing Limits and Final Answer

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Analyzing the Symmetry of the Function

The Bridge

The Integral Transformation

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Animated Solution for Mathematics - Definite Integration: 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-

Parity of $f (x)$

Substituting to find $f (- x)$

$f (x)$ is an Odd Function

Analyzing $f (x + 5) = g (x)$

Substituting $- x$

Final Relation between $f$ and $g$

Substituting $- g (u)$

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