The value of -13t2cos2t u- 3t-3-2 dt- is where u-t-dudt consider- -3-14

∫_ 1^3t^2cos2tu' ( 3t 3π2 )dt=∫_ 1^3t^2cos2tδ ( 3t 3π2 )dt u'(t)=ddtu(t)=δ (t) =13∫_ 1^3t^2cos2t.δ ( t π2 )dt [ ∵δ ( 3t 3π2 )=δ ( 3 ( t π2 ) )=13δ ( t π2 ) ...

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Byju's Answer

Standard XII

Mathematics

Square Root of a Complex Number

The value of ...

Question

The value of $\int_{- 1}^{3} t^{2} c o s 2 t u^{'} (3 t - \frac{3 π}{2}) d t,$ is ________ where $u^{'} (t) = \frac{d u}{d t}$ (consider $π = 3.14$ )
-0.8216

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Solution

The correct option is A -0.8216
$\int_{- 1}^{3} t^{2} c o s 2 t u^{'} (3 t - \frac{3 π}{2}) d t = \int_{- 1}^{3} t^{2} c o s 2 t δ (3 t - \frac{3 π}{2}) d t$

$u^{'} (t) = \frac{d}{d t} u (t) = δ (t)$

$= \frac{1}{3} \int_{- 1}^{3} t^{2} c o s 2 t . δ (t - \frac{π}{2}) d t$

$[∵ δ (3 t - 3 \frac{π}{2}) = δ (3 (t - \frac{π}{2})) = \frac{1}{3} δ (t - \frac{π}{2})]$

$= \frac{1}{3} \times {[t^{2} c o s 2 t]}_{t = \frac{π}{2}}^{2}$

$= \frac{1}{3} \times {(\frac{π}{2})}^{2} c o s (2 \times \frac{π}{2})$

$= \frac{1}{3} \times (1.57)^{2} \times c o s π$

$= \frac{2.465}{3} = - 0.8216$

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The value of ∫3−1t2cos2t u′(3t−3π2)dt, is ________ where u′(t)=dudt (considerπ=3.14)-0.8216

The correct option is A -0.8216∫3−1t2cos2tu′(3t−3π2)dt=∫3−1t2cos2tδ(3t−3π2)dt u′(t)=ddtu(t)=δ(t) =13∫3−1t2cos2t.δ(t−π2)dt [∵δ(3t−3π2)=δ(3(t−π2))=13δ(t−π2)] =13×[t2cos2t]t=π2 =13×(π2)2cos(2×π2) =13×(1.57)2×cosπ =2.4653=−0.8216

The value of $\int_{- 1}^{3} t^{2} c o s 2 t u^{'} (3 t - \frac{3 π}{2}) d t,$ is ________ where $u^{'} (t) = \frac{d u}{d t}$ (consider $π = 3.14$ )
-0.8216