Let f, g : (0, ∞) → R be two functions defined by f(x) = ∫(-x to x) (|t| - t²) e^(-t²) dt and g(x) =

Q: Why can't we just integrate $f(x)$ and $g(x)$ directly? They look like standard integrals.

That is a great question! While they might look standard, the combination of absolute values, polynomial terms, and exponential functions makes direct integration extremely tedious and prone to error. In JEE Advanced problems, when you see complex integral definitions, it is often a signal to look for properties like symmetry or to use the Leibniz Rule to differentiate the functions first. It is a strategic choice to simplify the problem before solving it.

Q: Wait, why do we need to find the constant $C$ after integrating the sum of the derivatives?

When you integrate $f'(x) + g'(x)$ to get $f(x) + g(x)$, you are finding the general form of the sum. The constant $C$ represents the family of all possible functions that share that derivative. To find the *specific* function that matches our definitions of $f(x)$ and $g(x)$, we must use the initial condition. Since $f(0) = 0$ and $g(0) = 0$, we know that at $x=0$, the sum must be zero. This allows us to pin down the exact value of $C$.

Q: Is it always safe to use the Leibniz Rule for these types of problems?

Yes, absolutely! The Leibniz Rule is a powerful tool in your arsenal. As long as the functions inside the integral are continuous and the limits are differentiable, you can differentiate under the integral sign. It is particularly useful when the integral itself is hard to evaluate but its derivative is simple. Always check if the derivative simplifies the expression before committing to it.

Question

Let $f, g : (0, \infty) 	o \mathbb{R}$ be two functions defined by $f(x) = \int_{-x}^{x} (|t| - t^2) e^{-t^2} dt$ and $g(x) = \int_{0}^{x^2} t^{1/2} e^{-t} dt$. Then the value of $9(f(\sqrt{\log_e 9}) + g(\sqrt{\log_e 9}))$ is equal to

Options:

(A) 6

(B) 9

(C) 8

(D) 10

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a complex mathematical landscape. You have two functions:
$$f(x) = \int_{-x}^{x} (|t| - t^2) e^{-t^2} dt$$
$$g(x) = \int_{0}^{x^2} t^{1/2} e^{-t} dt$$

At first glance, they look intimidating. But in the world of JEE Advanced, intimidation is just a mask for elegance. Let us peel back that mask together.

### Phase 1: The Symmetry Insight

Your first instinct when you see an integral with symmetric limits like $[-x, x]$ should always be to check for symmetry. Let us examine the integrand of $f(x)$, which we will call $h(t) = (|t| - t^2) e^{-t^2}$.

If you replace $t$ with $-t$, the absolute value $|-t|$ remains $|t|$, and $(-t)^2$ remains $t^2$. The exponential term $e^{-(-t)^2}$ is also unchanged. Thus, $h(-t) = h(t)$, making it an **even function**.

This is a gift! It allows us to rewrite $f(x)$ as:
$$f(x) = 2 \int_{0}^{x} (t - t^2) e^{-t^2} dt$$

Now, the absolute value is gone because $t$ is positive, and our integral is much cleaner.

### Phase 2: The Leibniz Strategy

Now, we face a choice. Should we integrate? No, that looks like a path into a thicket of complexity. Instead, let us use the **Newton-Leibniz Rule** to differentiate.

For $f(x)$, the derivative is:
$$f'(x) = 2(x - x^2) e^{-x^2}$$

For $g(x)$, we apply the rule to the upper limit $x^2$. The derivative becomes:
$$g'(x) = (x^2)^{1/2} e^{-x^2} \cdot \frac{d}{dx}(x^2) = x e^{-x^2} \cdot 2x = 2x^2 e^{-x^2}$$

### Phase 3: The Magical Cancellation

Now, watch closely. This is the moment where the problem reveals its true, elegant soul. When we add the derivatives:
$$f'(x) + g'(x) = (2x - 2x^2) e^{-x^2} + 2x^2 e^{-x^2}$$

The term $-2x^2 e^{-x^2}$ and $+2x^2 e^{-x^2}$ cancel out perfectly! We are left with a simple, beautiful expression:
$$f'(x) + g'(x) = 2x e^{-x^2}$$

### Phase 4: The Final Integration

We have the derivative of the sum, so we integrate it to find the sum itself:
$$f(x) + g(x) = \int 2x e^{-x^2} dx$$

Using the substitution $u = x^2$, where $du = 2x dx$, the integral becomes:
$$\int e^{-u} du = -e^{-u} + C = -e^{-x^2} + C$$

To find $C$, we use the fact that at $x=0$, both $f(0)$ and $g(0)$ are zero. Thus, $0 = -e^0 + C$, which means $C = 1$. Our sum is:
$$f(x) + g(x) = 1 - e^{-x^2}$$

### Phase 5: The Victory

Finally, we evaluate this at $x = \sqrt{\log_e 9}$. The square root and the square in the exponent cancel out, leaving us with:
$$1 - e^{-\log_e 9} = 1 - \frac{1}{9} = \frac{8}{9}$$

The question asks for $9(f+g)$, so:
$$9 	imes \frac{8}{9} = 8$$

We have arrived at the answer, not by brute force, but by understanding the structure of the problem. The final result is **8**. Keep this mindset, and no JEE problem will ever be too daunting.

Animated Solution for Mathematics - Definite Integration: Let $f, g : (0, \infty) \to R$ be two functions defined by $f (x) = \int_{- x}^{x} (∣ t ∣ - t^{2}) e^{- t^{2}} d t$ and $g (x) = \int_{0}^{x^{2}} t^{1/2} e^{- t} d t$ . Then the value of $9 (f (lo g_{e} 9) + g (lo g_{e} 9))$ is equal to

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Visualized Solution

Analyze the Symmetry of $f (x)$

Simplify $f (x)$ for $t \geq 0$

Differentiate $f (x)$ using Leibniz Rule

Differentiate $g (x)$ using Leibniz Rule

Sum the Derivatives $f^{'} (x) + g^{'} (x)$

Integrate to find $f (x) + g (x)$

Determine the Constant $C$

Evaluate at $x = lo g_{e} 9$

Final Calculation

The Sigma Insight: Newton-Leibniz & Reduction Formulas

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Phase 4

Phase 5

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Visualized Solution

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

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

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

Differentiate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g(x) using Leibniz Rule

Sum the Derivatives .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f′(x)+g′(x)

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

Determine the Constant .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C

Evaluate at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=loge​9​