If f(x)=∫0x(t3+2t2+3) dt\displaystyle {f{{\left({x}\right)}}}={\int_{{0}}^{{x}}}{\left({t}^{{3}}+{2}{t}^{{2}}+{3}\right)}\ {\left.{d}{t}\right.}f(x)=∫0x(t3+2t2+3) dt then f′′(x)=\displaystyle {f}{''}{\left({x}\right)}=f′′(x)= Preview Question 6
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