Integral Calculus> options - 1/24 17/168 1/7 None of these...

options\t- 1/2417/168\t1/7\tNone of these

Aditya Kartikeya , 10 Years ago

Grade 10

1 Answers

Jitender Singh

Last Activity: 10 Years ago

Ans:
Hello Student,
Please find answer to your question below

f(x) is continuous for all x.
$\int_{0}^{x}f(t)dt = \int_{1}^{x}t^2f(t)dt + \frac{x^{16}}{8}+\frac{x^6}{3}+a$ ….....(1)
Apply Newton – Leibniz to differentiate integral.
Differentiate both sided, we have
$f(x) = -x^2f(x) + \frac{16x^{15}}{8}+\frac{6x^5}{3}$
$f(x) = -x^2f(x) + 2x^{15}+2x^5$
$f(x) (1+x^2) = 2x^{15}+2x^5$
$f(x) = \frac{2x^{15}+2x^5}{1+x^2}$
$f(t) = \frac{2t^{15}+2t^5}{1+t^2}$
$\int_{0}^{x}f(t)dt = \int_{0}^{x}\frac{2t^{15}+2t^5}{1+t^2}dt$
$\int_{0}^{x}f(t)dt = \int_{0}^{x}\frac{2(t^{15}+t^5)}{1+t^2}dt$
For the integrand, do long devision
$\frac{t^{15}+t^5}{1+t^2}\approx t^{13}-t^{11}+t^9-t^7+t^5$
$\int_{0}^{x}f(t)dt = \int_{0}^{x}2( t^{13}-t^{11}+t^9-t^7+t^5)dt$
$\int_{0}^{x}f(t)dt = [2(\frac{t^{14}}{14}-\frac{t^{12}}{12}+\frac{t^{10}}{10}-\frac{t^8}{8}+\frac{t^6}{6})]_{0}^{x}$
$\int_{0}^{x}f(t)dt = [2(\frac{x^{14}}{14}-\frac{x^{12}}{12}+\frac{x^{10}}{10}-\frac{x^8}{8}+\frac{x^6}{6})]$
Similarly, we have
$\int t^2f(t)dt = \int \frac{2t^2(t^{15}+t^5)}{1+t^2}dt$
Doing the long division of integrand, we have
$\int t^2f(t)dt = \int [2(t^{15}-t^{13}+t^{11}-t^{9}+t^7)]dt$
$\int t^2f(t)dt =[2(\frac{t^{16}}{16}-\frac{t^{14}}{14}+\frac{t^{12}}{12}-\frac{t^{10}}{10}+\frac{t^{8}}{8})]+constant$
Putting these values in (1), you get the value of ‘a’.