Question

f(x) = integrate t ^ 5 dt from 0 to x ^ 3 * then; = Box; f^ prime (x)=

Answer 1

Given the function f(x) as shown below.

$f(x)=\int_0^{x3}t^5dt$

Simplify the integral

$\int_0^{x3}t^5dt=\lbrack\frac{t^{5+1}}{5+1}+C\rbrack_0^{x3}$$\lbrack\frac{t^{5+1}}{5+1}+C\rbrack_0^{x3}=\lbrack\frac{t^6}{6}+C\rbrack_0^{x3}$$(\frac{(x^3)^6}{6})-\frac{0^6}{6}$$\therefore f(x)=\frac{x^{18}}{6}$

Thus, the derivative of f(x) is:

$f^{\prime}(x)=\frac{18\times x^{18-1}}{6}=\frac{18x^{17}}{6}=3x^{17}$

Final answer

$f^{\prime}(x)=3x^{17}$