If ∫ x⁵ e⁻ˣ² dx = g(x) e⁻ˣ² + c, where c is a constant of integration, then g(-1) is equal to : (1)

Q: Why did we choose $t = x^2$ instead of $u = x^5$?

Great question! If you choose $u = x^5$, the derivative is $5x^4$, which doesn't help simplify the $e^{-x^2}$ term. Substitution is about finding a relationship between the variable and its derivative that simplifies the *entire* integrand. Here, $x^2$ is the 'inner' function of the exponential, and its derivative $2x$ is present in the $x^5$ term (as $x^4 \cdot x$). It's a perfect match!

Q: Why do we use Integration by Parts twice?

Think of IBP as a 'power reduction' tool for polynomials multiplied by exponentials. Each time you apply IBP, the power of the polynomial ($t^2 \to t^1 \to t^0$) decreases. We keep going until the polynomial part is just a constant, which makes the final integration trivial.

Q: Is there a shortcut for $\int t^n e^{-t} dt$?

Yes, there is! It's related to the Gamma function, but for JEE Advanced, mastering the IBP process is crucial. It builds the 'mathematical muscle' you need for more complex problems. Don't look for shortcuts until you've mastered the path!

Question

If $\int x^5 e^{-x^2} dx = g(x) e^{-x^2} + c$, where $c$ is a constant of integration, then $g(-1)$ is equal to :

Options:

(A) -5/2

(B) 1

(C) -1/2

(D) -1

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, aspiring mathematician! Today, we are going to embark on a journey through a beautiful integral that often appears in JEE Advanced examinations. We are tasked with evaluating the integral:

$$I = \int x^5 e^{-x^2} dx$$

Our goal is to find the function $g(x)$ such that the integral takes the form $g(x) e^{-x^2} + c$.

### The Art of Substitution

Look closely at the exponent of the exponential function: $-x^2$. In calculus, whenever you see a function and its derivative within the same integral, your mind should immediately jump to substitution.

We know that the derivative of $x^2$ is $2x$. Our integrand contains $x^5$, which we can cleverly rewrite as $x^4 \cdot x$. This $x$ is exactly what we need for our $dx$ term.

Let us set $t = x^2$. Differentiating both sides, we get $dt = 2x dx$, which implies $x dx = \frac{dt}{2}$.

Substituting these into our integral, we transform the expression:

$$I = \frac{1}{2} \int t^2 e^{-t} dt$$

### The Iterative Power of IBP

We now have a product of an algebraic function, $t^2$, and an exponential function, $e^{-t}$. This is the classic setup for Integration by Parts (IBP). Using the ILATE rule, we choose $u = t^2$ and $dv = e^{-t} dt$.

Applying the IBP formula $\int u dv = uv - \int v du$, we obtain:

$$\frac{1}{2} \left[ t^2 (-e^{-t}) - \int (2t)(-e^{-t}) dt \right] = \frac{1}{2} \left[ -t^2 e^{-t} + 2 \int t e^{-t} dt \right]$$

Notice that the power of $t$ has dropped from $2$ to $1$. We apply IBP one more time to the remaining integral $\int t e^{-t} dt$:

$$\int t e^{-t} dt = t(-e^{-t}) - \int (1)(-e^{-t}) dt = -t e^{-t} - e^{-t}$$

### The Final Assembly

Now, let's bring it all together by substituting our result back into the main equation:

$$I = \frac{1}{2} \left[ -t^2 e^{-t} + 2(-t e^{-t} - e^{-t}) \right] + c$$

Distributing the terms, we get:

$$I = -e^{-t} \left( \frac{t^2}{2} + t + 1 \right) + c$$

Substituting $t = x^2$ back into the expression to return to our original variable $x$:

$$I = -e^{-x^2} \left( \frac{x^4}{2} + x^2 + 1 \right) + c$$

By comparing this with the given form $g(x) e^{-x^2} + c$, we identify:

$$g(x) = -\left( \frac{x^4}{2} + x^2 + 1 \right)$$

### Final Calculation

The final step is to evaluate $g(-1)$. Plugging in $x = -1$:

$$g(-1) = -\left( \frac{(-1)^4}{2} + (-1)^2 + 1 \right) = -\left( \frac{1}{2} + 1 + 1 \right)$$

**The final result is:**

$$g(-1) = -\frac{5}{2}$$

Animated Solution for Mathematics - Indefinite Integration: If $\int x^{5} e^{- x^{2}} d x = g (x) e^{- x^{2}} + c$ , where $c$ is a constant of integration, then $g (- 1)$ is equal to :

Select Answer:

Visualized Solution

Understanding the Goal

Choosing the Substitution

Rewriting the Integral

First Integration by Parts

Simplifying the First IBP

Second Integration by Parts

Completing the Integration

Distributing and Factoring

Back Substitution

Identifying $g (x)$

Evaluating $g (- 1)$

Final Conclusion

The Sigma Insight: Integration by Substitution

Analyzing the Setup

The Art of Substitution

The Iterative Power of IBP

The Final Assembly

Final Calculation

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Animated Solution for Mathematics - Indefinite Integration: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If ∫x5e−x2dx=g(x)e−x2+c, where c is a constant of integration, then g(−1) is equal to :