Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSINTEGRALS3 Marks
Question
Evaluate the definite integral in Exercise:$\int_{1}^{2}(4\text{x}^{3}-5\text{x}^{2}+6\text{x}+9)\text{dx}$
✓
Answer
$\text{Let}=\int\limits_{1}^{2}(4\text{x}^{3}-5\text{x}^{2}+6\text{x}+9)\text{dx}$$\int(4\text{x}^{3}-5\text{x}^{2}+6\text{x}+9)\text{dx}=4\bigg(\frac{\text{x}^{4}}{4}\bigg)-5\bigg(\frac{\text{x}^{3}}{3}\bigg)+6\bigg(\frac{\text{x}^{2}}{2}\bigg)+9\text{(x)}$
$=\text{x}^{4}-\frac{5\text{x}^{3}}{3}+3\text{x}^{2}+9\text{x}=\text{F}\text{(x)}$
By second fundamental theorem of calculus, we obtain
$\text{I}=\text{F}(2)-\text{F}(1)$
$\text{I}=\Bigg\{2^{4}-\frac{5.(2)^{3}}{3}+3(2)^{2}+9(2)\Bigg\}-\bigg\{(1)^{4}-\frac{5(1)^{3}}{3}+3(1)^{2}+9(1)\bigg\}$
$=\bigg(16-\frac{40}{3}+12+18\bigg)-\bigg(1-\frac{5}3{+3+9}\bigg)$
$=16-\frac{40}{3}+12+18-1+\frac{5}{3}-3-9$
$=33-\frac{35}{3}$
$=\frac{99-35}{3}$
$=\frac{64}{3}$
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