11YAAR SACCHI
CALCULAS HAIN!!!!!!!!!!!!!!!!!!!!!!!!!
i personally consider that these type of problems have to be done by special way
because this question is useless without the options
iss pe tipaani karne ke baad [3]
let us think of this question[12][12][12]
11WAISE EK QUES ISKE BAARE MEIN
HOW WOULD α AND β COME IN THE PROBLEM
THERE IS NO MENTION ABOUT IT
JUST THEIR RANGES R MENTIONED
MOREOVER I THINK TAPAN THE WAY U PROCEED IS THE ONLY 1 WHICH CAME TO MY MIND!!!!!!!!!
66Use the substitution t=x^4. Then, \mathrm{d}t=4x^3\ \mathrm{d}x. Hence, the given integral becomes, (after suitably scaling the limits)
\int_0^24x^3f(x^4)\ \mathrm{d}x
Now, break in to two integrals:
I=\int_0^24x^3f(x^4)\ \mathrm{d}x=\int_0^14x^3f(x^4)\ \mathrm{d}x + \int_1^24x^3f(x^4)\ \mathrm{d}x
By mean value theorem, there exists some \alpha satisfying 0<\alpha<1 and \beta satisfying 1<\beta <2 such that
\int_0^1 4x^3 f(x^4)\ \mathrm{d}x=4\alpha^3f(\alpha^4)
and
\int_1^2 4x^3 f(x^4)\ \mathrm{d}x=4\beta^3f(\beta^4)
Accordingly, we get
\int_0^{16}f(t)\mathrm{d}t=4\left(\alpha^3f(\alpha^4)+\beta^3f(\beta^4)\right)
for 0<α<1 and 1<β<2. So the answer is (B).
1GR8 job [1]
but sir i didnt understand the
2nd step like how u wrote the f(x4) PART ..pls explain
