How Cheenta works to ensure student success?
Explore the Back-Story

Definite Integral | IIT JAM 2018 | Problem 4

Try this beautiful problem from IIT JAM 2018 which requires knowledge of the properties of Definite integral.

Properties of Definite Integral -IIT JAM2018 (Problem 4)


Let a be positive real number. If f is a continuous and even function defined on the interval [-a,a], then \displaystyle\int_{-a}^a \frac{f(x)}{1+e^x} \mathrm d x is equal to :-

  • \displaystyle\int_0^a f(x) \mathrm d x
  • 2\displaystyle\int_0^a \frac{f(x)}{1+e^x}\mathrm d x
  • 2\displaystyle\int_0^a f(x) \mathrm d x
  • 2a\displaystyle\int_0^a \frac{f(x)}{1+e^x}\mathrm d x

Key Concepts


Definite Integral

Properties of definite Integral

Even function / Odd function

Check the Answer


Answer: \displaystyle\int_0^a f(x) \mathrm d x

IIT JAM 2018, Problem 4

Definite and Integral calculus : R Courant

Try with Hints


In this first I will give you the properties we need to solve this problem :

Property 1 : \displaystyle\int_a^b f(x) \mathrm d x =  \displaystyle\int_a^b f(a+b-x) \mathrm d x

[Where f is continuous on [a,b]]

Property 2 : If f is an even function i.e., f(x)=f(-x) then

\displaystyle\int_{-a}^{a} f(x) \mathrm d x = 2 \displaystyle\int_{0}^{a} f(x) \mathrm d x

Can you drive it from here !!!! Give it a try !!!

Let I=\displaystyle\int_{-a}^a \frac{f(x)}{1+e^x} \mathrm d x \quad \ldots (i)

\Rightarrow I=  \displaystyle\int_{-a}^a \frac{f(a-a-x)}{1+e^{(a-a-x)}} \mathrm d x

[Since, f is continuous then \displaystyle\int_{a}^b  f(x) \mathrm{d}x =  \displaystyle\int_{a}^b  f(a+b-x) \mathrm{d} x]

\Rightarrow I=  \displaystyle\int_{-a}^a  \frac{f(-x)}{1+e^{-x}} \mathrm d x

\Rightarrow I=  \displaystyle\int_{-a}^a  \frac{f(x)}{1+\frac{1}{e^x}} \mathrm d x [Since f(x) is even]

\Rightarrow I= \displaystyle\int_{-a}^a  \frac{e^x.f(x)}{1+e^{x}} \mathrm d x \quad \ldots (ii)

Adding (i) and (ii) we can get some interesting result !!!

Adding (i) and (ii) we get ,

2I= \displaystyle\int_{-a}^a  \frac{f(x)}{1+e^{x}} \mathrm d x +  \displaystyle\int_{-a}^a  \frac{e^x .f(x)}{1+e^{x}} \mathrm d x

\Rightarrow 2I = \displaystyle\int_{-a}^a  \frac{[f(x)+e^x.f(x)]}{1+e^{x}} \mathrm d x

\Rightarrow 2I = \displaystyle\int_{-a}^a  \frac{f(x)[1+e^x]}{[1+e^{x}]}

\Rightarrow 2I=  \displaystyle\int_{-a}^a  f(x) \mathrm d x

\Rightarrow 2I = 2\displaystyle\int_0^a f(x) \mathrm d x [Since f(x) is even ]

\Rightarrow I = \displaystyle\int_0^a f(x) \mathrm d x [ANS]

Subscribe to Cheenta at Youtube


Try this beautiful problem from IIT JAM 2018 which requires knowledge of the properties of Definite integral.

Properties of Definite Integral -IIT JAM2018 (Problem 4)


Let a be positive real number. If f is a continuous and even function defined on the interval [-a,a], then \displaystyle\int_{-a}^a \frac{f(x)}{1+e^x} \mathrm d x is equal to :-

  • \displaystyle\int_0^a f(x) \mathrm d x
  • 2\displaystyle\int_0^a \frac{f(x)}{1+e^x}\mathrm d x
  • 2\displaystyle\int_0^a f(x) \mathrm d x
  • 2a\displaystyle\int_0^a \frac{f(x)}{1+e^x}\mathrm d x

Key Concepts


Definite Integral

Properties of definite Integral

Even function / Odd function

Check the Answer


Answer: \displaystyle\int_0^a f(x) \mathrm d x

IIT JAM 2018, Problem 4

Definite and Integral calculus : R Courant

Try with Hints


In this first I will give you the properties we need to solve this problem :

Property 1 : \displaystyle\int_a^b f(x) \mathrm d x =  \displaystyle\int_a^b f(a+b-x) \mathrm d x

[Where f is continuous on [a,b]]

Property 2 : If f is an even function i.e., f(x)=f(-x) then

\displaystyle\int_{-a}^{a} f(x) \mathrm d x = 2 \displaystyle\int_{0}^{a} f(x) \mathrm d x

Can you drive it from here !!!! Give it a try !!!

Let I=\displaystyle\int_{-a}^a \frac{f(x)}{1+e^x} \mathrm d x \quad \ldots (i)

\Rightarrow I=  \displaystyle\int_{-a}^a \frac{f(a-a-x)}{1+e^{(a-a-x)}} \mathrm d x

[Since, f is continuous then \displaystyle\int_{a}^b  f(x) \mathrm{d}x =  \displaystyle\int_{a}^b  f(a+b-x) \mathrm{d} x]

\Rightarrow I=  \displaystyle\int_{-a}^a  \frac{f(-x)}{1+e^{-x}} \mathrm d x

\Rightarrow I=  \displaystyle\int_{-a}^a  \frac{f(x)}{1+\frac{1}{e^x}} \mathrm d x [Since f(x) is even]

\Rightarrow I= \displaystyle\int_{-a}^a  \frac{e^x.f(x)}{1+e^{x}} \mathrm d x \quad \ldots (ii)

Adding (i) and (ii) we can get some interesting result !!!

Adding (i) and (ii) we get ,

2I= \displaystyle\int_{-a}^a  \frac{f(x)}{1+e^{x}} \mathrm d x +  \displaystyle\int_{-a}^a  \frac{e^x .f(x)}{1+e^{x}} \mathrm d x

\Rightarrow 2I = \displaystyle\int_{-a}^a  \frac{[f(x)+e^x.f(x)]}{1+e^{x}} \mathrm d x

\Rightarrow 2I = \displaystyle\int_{-a}^a  \frac{f(x)[1+e^x]}{[1+e^{x}]}

\Rightarrow 2I=  \displaystyle\int_{-a}^a  f(x) \mathrm d x

\Rightarrow 2I = 2\displaystyle\int_0^a f(x) \mathrm d x [Since f(x) is even ]

\Rightarrow I = \displaystyle\int_0^a f(x) \mathrm d x [ANS]

Subscribe to Cheenta at Youtube


Leave a Reply

Your email address will not be published. Required fields are marked *

This site uses Akismet to reduce spam. Learn how your comment data is processed.

Knowledge Partner

Cheenta is a knowledge partner of Aditya Birla Education Academy
Cheenta

Cheenta Academy

Aditya Birla Education Academy

Aditya Birla Education Academy

Cheenta. Passion for Mathematics

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.
JOIN TRIAL
support@cheenta.com
Menu
Trial
Whatsapp
magic-wandrockethighlight