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Aditya Kartikeya , 10 Years ago
Grade 10
anser 1 Answers
Jitender Singh

Last Activity: 10 Years ago

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


\int_{0}^{x}f(t)dt = \int_{1}^{x}t^2f(t)dt + \frac{x^{16}}{8}+\frac{x^6}{3}+a............(1)
f(x) is continuous for all x.

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’.

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