Question-223631

Question Number 223631 by MathematicalUser2357 last updated on 01/Aug/25

Commented by MathematicalUser2357 last updated on 01/Aug/25

$$\mathrm{Find}\:{F}\left({x}\right) \\ $$

Answered by Frix last updated on 01/Aug/25

∫(x^2 +4x)dx=(x^3 /3)+2x^2 +C=F(x) F(1)=(1/3)+2+C=(7/3)+C=−2 ⇒ C=−((13)/3) F(x)=(x^3 /3)+2x^2 −((13)/3)

$$\int\left({x}^{\mathrm{2}} +\mathrm{4}{x}\right){dx}=\frac{{x}^{\mathrm{3}} }{\mathrm{3}}+\mathrm{2}{x}^{\mathrm{2}} +{C}={F}\left({x}\right) \\ $$$${F}\left(\mathrm{1}\right)=\frac{\mathrm{1}}{\mathrm{3}}+\mathrm{2}+{C}=\frac{\mathrm{7}}{\mathrm{3}}+{C}=−\mathrm{2}\:\Rightarrow\:{C}=−\frac{\mathrm{13}}{\mathrm{3}} \\ $$$${F}\left({x}\right)=\frac{{x}^{\mathrm{3}} }{\mathrm{3}}+\mathrm{2}{x}^{\mathrm{2}} −\frac{\mathrm{13}}{\mathrm{3}} \\ $$

Answered by fantastic last updated on 01/Aug/25

F^( ′) (x)=x^2 +4x So F(x)=∫F^( ′) (x)=∫x^2 +4x=(x^(2+1) /(2+1))+2x^2 +C So F(x)=(x^3 /3)+2x^2 +C now F(1)=−2 so (1^3 /3)+2×1^2 +C=−2 or (1/3)+C=−2−2=−4 So C=−4−(1/3)= −((13)/3) So F(x)=(x^3 /3)+2x^2 −((13)/3)✓

$${F}^{\:'} \left({x}\right)={x}^{\mathrm{2}} +\mathrm{4}{x} \\ $$$${So}\:{F}\left({x}\right)=\int{F}^{\:'} \left({x}\right)=\int{x}^{\mathrm{2}} +\mathrm{4}{x}=\frac{{x}^{\mathrm{2}+\mathrm{1}} }{\mathrm{2}+\mathrm{1}}+\mathrm{2}{x}^{\mathrm{2}} +{C} \\ $$$${So}\:{F}\left({x}\right)=\frac{{x}^{\mathrm{3}} }{\mathrm{3}}+\mathrm{2}{x}^{\mathrm{2}} +{C} \\ $$$${now}\:\:{F}\left(\mathrm{1}\right)=−\mathrm{2} \\ $$$${so}\:\frac{\mathrm{1}^{\mathrm{3}} }{\mathrm{3}}+\mathrm{2}×\mathrm{1}^{\mathrm{2}} +{C}=−\mathrm{2} \\ $$$${or}\:\frac{\mathrm{1}}{\mathrm{3}}+{C}=−\mathrm{2}−\mathrm{2}=−\mathrm{4} \\ $$$${So}\:{C}=−\mathrm{4}−\frac{\mathrm{1}}{\mathrm{3}}=\:−\frac{\mathrm{13}}{\mathrm{3}} \\ $$$${So}\:{F}\left({x}\right)=\frac{{x}^{\mathrm{3}} }{\mathrm{3}}+\mathrm{2}{x}^{\mathrm{2}} −\frac{\mathrm{13}}{\mathrm{3}}\checkmark \\ $$

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