Question

If f(x) is continuous ∀ x∈R and

$\int_{0}^{x}{f(t)dt = \int_{x}^{1}}$t²f(t) dt + $\frac{x^{16}}{8} + \frac{x^{6}}{3}$+ a then value of a is equal to

• A. $- \frac{1}{24}$
• B. $\frac{17}{168}$
• C. $\frac{1}{7}$
• D. $- \frac{167}{840}$

Answer 1

Answer: (B)

$\frac{17}{168}$

Answer 2

diff. w.r.t. x to get

f ′(x) = – x²f(x) + 2x¹⁵ + 2x⁵

or $\frac{dy}{dx}$+ x²y = 2x¹⁵ + 2x⁵

⇒ y. $e^{x^{3}/3}$= $2\int_{}^{}e^{x^{3}/3}$ (x¹⁵ + x⁵)dx