Integration-3

Calculus Level 4

$\large \dfrac{ e^{2x} (2x-5)}{(2x-3)^3}$

If an antiderivative the expression above is in the form of

$\dfrac ab \cdot \dfrac{ e^{2x}}{(2x-3)^c} \; ,$

where $a,b$ and $c$ are positive integers with $a,b$ coprime, find $a\times b+c$ .

The answer is 4.

1 solution

Rishabh Jain
Mar 14, 2016

$2x=t$ $\large \dfrac 12 \int e^{t}\left({\frac{1}{(t-3)^2}}-\dfrac{2}{(t-3)^3}\right)dt$ This is of the form $\int e^x(f(x)+f'(x))dx=e^xf(x)$ $\therefore \int\frac{(2 x-5)(e^x)^2}{(2 x-3)^3}=\dfrac{1}{2}\left(\frac{(e^x)^2}{(2 x-3)^2}\right)$ Hence $1\times 2+2=4$ .

I faced this question in today's Maths board paper .. Where did you get it from??

Nice solution.

Correction: Make it $2x=t$ .

Harsh Khatri - 5 years, 3 months ago

Aah.. Right... Thanks :-)

Rishabh Jain - 5 years, 3 months ago

Same solution! Easy one!

Adarsh Kumar - 5 years, 3 months ago

Even I faced this question in today's board paper.

Akshay Sharma - 5 years, 3 months ago

Have you lost some marks due to length of paper because many have lost marks because this year too paper was lengthy... :-)

Rishabh Jain - 5 years, 3 months ago

yup! paper was bit lengthy and I was not expecting that 6 mark question from relation and functions seriously

Akshay Sharma - 5 years, 3 months ago

@Akshay Sharma – Some questions were really good while some were just there to recover time ... That LPP problem took my most of the time ... It was ridiculous..

Rishabh Jain - 5 years, 3 months ago

@Rishabh Jain – yeah and I wasn't mentally prepared for a question on binary operation .

Akshay Sharma - 5 years, 3 months ago

@Akshay Sharma – True.. I read it today morning before paper :-) .. Some of my friends too left that topic as it was unimportant for JEE Adv.

Rishabh Jain - 5 years, 3 months ago

$\int{(x+3)\sqrt(3-4 x-x^2)}dx$ what about this question.I just lost my cool solving this

Akshay Sharma - 5 years, 3 months ago

$\int(x+3)\sqrt{(3-4 x-x^2)}$ $=\dfrac{-1}{2} \left(\int(-2x-4)-2\sqrt{3-4 x-x^2}\right)$ $\underbrace{\dfrac{-1}{2}\left((-2x-4)\sqrt{3-4 x-x^2}\right)dx}_{I_1}+\underbrace{\int\sqrt{7-(x+2)^2}dx}_{I_2}$ $I_1=\int (\sqrt{3-4 x-x^2})d(\sqrt{3-4 x-x^2})$ (Or make a substitution $\sqrt{3-4 x-x^2}=t$

While for $I_2$ apply the formula for $\int\sqrt{a^2-x^2}dx$

It took two pages of my answer sheet. I solved it in very last since I couldn't recollect the formula for $\int \sqrt{a^2-x^2}dx$ ........

Rishabh Jain - 5 years, 3 months ago

Thanks ,nice solution

Akshay Sharma - 5 years, 3 months ago

The same goes for me. I am rarely able to recollect the formulae for $\int \sqrt{a^2\pm x^2}dx, \int \frac{dx}{a^2 \pm x^2}$ , etc.

I always end up using trigonometric substitutions for these integrals.

Harsh Khatri - 5 years, 3 months ago

@Harsh Khatri – Exactly the same story with me .... I also end up using trig. Subst.. :-)

Rishabh Jain - 5 years, 3 months ago

Integration-3

The answer is 4.

1 solution

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