1

I found the following problem in JEE-Advanced (2014) question paper:

$$\int_{-2}^{2} \frac{3x^2}{1+e^x}\mathrm dx$$

At present, I don't know how to approach the problem. So, a few good hints would be appreciated.

(It seems there is no closed form for the antiderivative of the function.)

BAI
  • 2,565
Partha Sarker
  • 547

3 Answers3

16

Let $u=-x$.

\begin{align} \int_{-2}^2\frac{3x^2}{1+e^x}dx&=\int_2^{-2}\frac{-3u^2}{1+e^{-u}}du\\ &=\int_{-2}^2\frac{3u^2}{1+e^{-u}}du\\ &=\int_{-2}^2\frac{3u^2e^u}{1+e^{u}}du\\ &=\int_{-2}^2\frac{3x^2e^x}{1+e^x}dx \end{align}

So

$$2\int_{-2}^2\frac{3x^2}{1+e^x}dx=\int_{-2}^2\frac{3x^2}{1+e^x}dx+\int_{-2}^2\frac{3x^2e^x}{1+e^x}dx=3\int_{-2}^2x^2dx$$

CY Aries
  • 23,393
  • 3
    Slick. (+1) ${}$ – Chris Jun 13 '17 at 01:15
  • I don't understand the 3rd line. – Partha Sarker Jun 13 '17 at 01:36
  • 1
    Do you mean $\int_2^{-2}\frac{-3u^2}{1+e^{-u}}du=\int_{-2}^2\frac{3x^2e^x}{1+e^x}dx$? Multiply both the denominator and numerator by $e^u$, then replace the dummy variables by $x$. – CY Aries Jun 13 '17 at 01:38
  • 1
    Sorry, I don't which part you are referring to. The 3rd line is "So". – CY Aries Jun 13 '17 at 01:42
  • but u=-x right? Then, it would change the whole expresseion to the previous one! – Partha Sarker Jun 13 '17 at 01:42
  • The previous line of "So" – Partha Sarker Jun 13 '17 at 01:44
  • 2
    damnn hella smart +1 – Saketh Malyala Jun 13 '17 at 01:45
  • @Mockingbird360 I just edited the proof to make it easier to read. Do you mean $\int_{-2}^2\frac{3u^2e^u}{1+e^{u}}du=\int_{-2}^2\frac{3x^2e^x}{1+e^x}dx$? It is to replace the dummy variable. – CY Aries Jun 13 '17 at 01:46
  • Yes, I meant that. but u=-x? what about that? – Partha Sarker Jun 13 '17 at 01:50
  • For definite integrals, variables sometimes are not really "variables" and can be replaced by other "variables". For example, $\int e^xdx$ and $\int e^udu$ are not the same, as the first one is $e^x+C_1$ and the second one is $e^x+C_2$. But $\int_1^2e^xdx$ and $\int_1^2e^udu$ are actually the same. Both of them are equal to $e^2-e$. $\int_{-2}^2\frac{3u^2e^u}{1+e^{u}}du$ and $ \int_{-2}^2\frac{3x^2e^x}{1+e^x}dx$ are the same, the "variables" $x$ and $u$ are called dummy variables. The value of the definite integrals does not depend on $u$ or $x$ – CY Aries Jun 13 '17 at 01:56
  • I just replace $u$ by $x$ as the integrals are the same after the replacement. This is not a substitution. – CY Aries Jun 13 '17 at 01:57
  • Very cool. Would you mind sharing the motivation for this idea? I graphed $\displaystyle \dfrac {3x^2}{1+x^2}$ and $\dfrac {3x^2}{1+x^2} \cdot e^x$ and indeed, the functions are reflections of each other with respect to the $x$ axis. But I would've never guessed it. – Ovi Jun 13 '17 at 01:59
  • @Ovi As the indefinite integral has closed form in terms of elementary functions, not much I can do. $u=-x$ is a natural choice of substitution since it keeps the upper and the lower limits. This skill is common in examination papers. – CY Aries Jun 13 '17 at 02:03
  • @CYAries Thanks for the response. But if you don't mind, how did you know the indefinite integral does not have a closed form in terms of elementary functions? When I saw the question title my first idea was to try integration by parts and hope to get a closed form. – Ovi Jun 13 '17 at 02:06
  • @Ovi I think it is hard to answer your question completely. You may refer to https://math.stackexchange.com/questions/155/how-can-you-prove-that-a-function-has-no-closed-form-integral. Many indefinite integrals has no closed form in elementary functions. When you tried all possible common skills without success, it is likely that the integral is one the those. – CY Aries Jun 13 '17 at 02:13
  • Thanks for the help, I will check out the link. – Ovi Jun 13 '17 at 03:31
1

Although the question has been answered in a similar way, I would like to throw some light on the general strategy for solving definite integrals of this sort. Let us consider $I=\int_a^b f(x) dx$. Substitute $u=a+b-x$. $$I=\int_a^bf(x)dx=-\int_b^af(a+b-x)dx=\int_a^bf(a+b-x)dx \tag1$$Note that this property is useful especially when $f(x)$ is a rational function whose denominator remains the same upon this substitution. Considering the integral the OP has asked. Let $$I=\int_{-2}^2\dfrac{3x^2}{1+e^x}dx=\int_{-2}^{2}\dfrac{3x^2e^x}{1+e^x}dx \implies I=\dfrac{x^3}{3}\mid_0^2=8$$ Note: This answer is written keeping the typology of JEE Definite Integration questions in mind. Definite Integration questions in the JEE usually requires one to make use of formula $(1)$.

Paras Khosla
  • 6,481
0

Here is my own method.

$$\int_{-2}^{2}\frac{3x^2}{1+e^x}\mathrm dx$$ $$=\int_{-2}^{0} \frac{3x^2}{1+e^x}\mathrm dx+ \int_{0}^{2}\frac{3x^2}{1+e^x}\mathrm dx $$

For $x\in[-2,0]$ the fraction $\dfrac{3x^2}{1+e^x}$ becomes $\dfrac{3x^2 e^{-x}}{1+e^{-x}}$

So, $\displaystyle\int_{-2}^{0} \frac{3x^2}{1+e^x}\mathrm dx $ becomes

$$\int_{0}^{2}\frac{3x^2 e^x}{1+e^x}\mathrm dx$$

Now the expression simplifies, $$\int_{0}^{2}\frac{3x^2e^x}{1+e^x} \mathrm dx+ \int_{0}^{2}\frac{3x^2}{1+e^x}\mathrm dx=\int_{0}^{2} 3x^2\mathrm dx $$ = 8; which is the answer in my key.

Paras Khosla
  • 6,481
Partha Sarker
  • 547