The function $$f(x)=$$∫$$1x$$[$$2(t$$−$$1)(t$$−$$2)3+3(t$$−$$1)2(t$$−$$2)2$$] dt has:

Question

The  function $$f(x)=$$∫$$1x$$[$$2(t$$−$$1)(t$$−$$2)3+3(t$$−$$1)2(t$$−$$2)2$$] dt      has:

Infinity Learn · Accepted Answer

$$f(x)=$$∫$$1x{2(t$$−$$1)(t$$−$$2)3+3(t$$−$$1)2(t$$−$$2)2}$$ dt       ⇒$$f1(x)=2(x$$−$$1)(x$$−$$2)3+3(x$$−$$1)2(x$$−$$2)2$$     For maximum (or) minimum $$f1(x)=0$$     ⇒$$2(x$$−$$1)(x$$−$$2)3+3(x$$−$$1)2(x$$−$$2)2=0$$     ⇒(x$$−$$1)(x$$−$$2)2{2(x$$−$$2)+3(x$$−$$1)}=0$$     ⇒$$(x$$−$$1)(x$$−$$2)2{5x$$−$$7}=0$$     $$x=1$$, $$2$$, $$7/5$$     now  $$f1(x)=(x$$−$$1)(x$$−$$2)2(5x$$−$$7)     ⇒$$f1(x)=(x$$−$$2)2(5x$$−$$7)+2(x$$−$$1)(x$$−$$2)(5x$$−$$3)+(x$$−$$1)(x$$−$$2)25$$........(2)     Putting $$x=1$$,$$x=2$$,$$x=7/5$$  $$is(2)$$ we get     $$f11(1)$$<$$0$$           $$f11(2)=0$$          $$f11(7/5)$$>$$0$$

The function $f (x) = \int_{1}^{x} [2 (t - 1) {(t - 2)}^{3} + 3 {(t - 1)}^{2} {(t - 2)}^{2}] d t$ has:

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The function f(x)=∫1x[2(t−1)(t−2)3+3(t−1)2(t−2)2] dt has:

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The function $f (x) = \int_{1}^{x} [2 (t - 1) {(t - 2)}^{3} + 3 {(t - 1)}^{2} {(t - 2)}^{2}] d t$ has: