Question
If $$f\left( x \right) = \int_0^x {\left( {{t^2} + 2t + 2} \right)dt,\,2 \leqslant x \leqslant 4,} $$ then :
A.
the maximum value of $$f\left( x \right)$$ is $$\frac{{136}}{3}$$
B.
the minimum value of $$f\left( x \right)$$ is 10
C.
the maximum value of $$f\left( x \right)$$ is 26
D.
none of these
Answer :
the maximum value of $$f\left( x \right)$$ is $$\frac{{136}}{3}$$
Solution :
$$\eqalign{
& f'\left( x \right) = {x^2} + 2x + 2 = {\left( {x + 1} \right)^2} + 1 > 0\,{\text{for all}}\,x \cr
& {\text{So,}}\,f\left( x \right)\,{\text{is m}}{\text{.i}}{\text{. in}}\,\left[ {2,\,4} \right] \cr
& \therefore \,\min f\left( x \right) = f\left( 2 \right){\text{ and }}\max f\left( x \right) = f\left( 4 \right) \cr
& \therefore \,\min f\left( x \right) = \int_0^2 {\left( {{t^2} + 2t + 2} \right)dt} = \left[ {\frac{{{t^3}}}{3} + {t^2} + 2t} \right]_0^2 = \frac{{32}}{3} \cr
& \therefore \,\max f\left( x \right) = \int_0^4 {\left( {{t^2} + 2t + 2} \right)dt} = \left[ {\frac{{{t^3}}}{3} + {t^2} + 2t} \right]_0^4 = \frac{{136}}{3} \cr} $$