Question

11. Evaluation of ſe? dt = x, then x will be equal to
Iron
[NCERT Pg. 48]
(1) Zero
(2) 3
(4) 27
(3) 9
tha​

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Value\:of\:x=9}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline{\bold{Given :}}} \\ \tt: \implies \int \limits_{0} ^{3} {t}^{2} dt = x \\ \\ \red{\underline{\bold{To \: Find :}}} \\ \tt : \implies Value \: of \: x = ?$

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt: \implies \int \limits_{0}^{3} {t}^{2} dt = x\\\\ \green{ \star} \tt \: \int {x}^{n} \: dx = \frac{ {x}^{n + 1} }{n + 1} \\ \\ \\ \tt: \implies x = \bigg[ \frac{ {t}^{3} }{3} \bigg]_{0}^{2} \\ \\ \tt: \implies x = \bigg[ \frac{ {3}^{3} }{3} - \frac{ {0}^{3} }{3} \bigg] \\ \\ \tt: \implies x = \frac{27}{3} \\ \\ \green{\tt: \implies x = 9}\\\\\green{\tt{\therefore{Value\:of\:x=9}}}$

Answer 2

SOLUTION

CORRECT QUESTION

TO CHOOSE THE CORRECT OPTION

Evaluation of

$\displaystyle \sf \int\limits_{0}^{3} {t}^{2} \, dt = x$

Then x will be

(1) Zero

(2) 3

(3) 9

(4) 27

EVALUATION

Here the given Integral is

$\displaystyle \sf \int\limits_{0}^{3} {t}^{2} \, dt = x$

On integration we get

$\displaystyle \sf \int\limits_{0}^{3} {t}^{2} \, dt = x$

$\displaystyle \sf \implies \: \bigg[ \frac{ {t}^{3} }{3} \bigg]_{0}^{3} = x$

$\displaystyle \sf \implies \: \bigg[ \frac{ {3}^{3} }{3} - \frac{ {0}^{3} }{3} \bigg] = x$

$\displaystyle \sf \implies \: \bigg[ \frac{27}{3} - \frac{0 }{3} \bigg] = x$

$\displaystyle \sf \implies 9 - 0= x$

$\displaystyle \sf \implies x = 9$

FINAL ANSWER

Hence the correct option is (C) 9

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