Question

find integral value of π/2.d theta with limit -4 to -1​

Answer 1

$\implies \displaystyle \sf\int\limits^{-1}_{-4} {\dfrac{\pi }{2} } \, d \theta$

$\implies\displaystyle \sf{\dfrac{\pi }{2} }\int\limits^{-1}_{-4} 1\, d \theta$

We know that :-

$\underline{\displaystyle \bf \boxed{\int\limits 1\, d x= x + c} }$

Therefore now :-

$\implies\displaystyle \sf{\dfrac{\pi }{2} }\int\limits^{-1}_{-4} 1\, d \theta = (\dfrac{\pi }{2} \times \theta)]^{-1} _{-4}$

Now we will substract lower limit from upper limit.

$\implies\displaystyle \sf (\dfrac{\pi }{2} \times \theta)]^{-1} _{-4}$

$\implies\displaystyle \sf (\dfrac{\pi }{2} \times - 1) - (\dfrac{\pi }{2} \times - 4)$

$\implies\displaystyle \sf ( - \dfrac{\pi }{2} ) - (\dfrac{ - 4\pi }{2} )$

$\implies\displaystyle \sf - \dfrac{\pi }{2} + \dfrac{ 4\pi }{2}$

$\implies\displaystyle \sf \dfrac{ 3\pi }{2}$

Answer :

$\displaystyle \sf\int\limits^{-1}_{-4} {\dfrac{\pi }{2} } \, d \theta = \bf \dfrac{ 3\pi }{2}$

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Learn more :-

∫ 1 dx = x + C

∫ sin x dx = – cos x + C

∫ cos x dx = sin x + C

∫ sec2 dx = tan x + C

∫ csc2 dx = -cot x + C

∫ sec x (tan x) dx = sec x + C

∫ csc x ( cot x) dx = – csc x + C

∫ (1/x) dx = ln |x| + C

∫ ex dx = ex+ C

∫ ax dx = (ax/ln a) + C

Answer 2

Answer:

The value of given integral under given limits is

$\frac{3\pi}{2}$

Explanation:

Given that,we have to find the integral value of 0.5πdθ with upper and lower limits -1 and -4 respectively

$I = \int _{ - 4 }^{ - 1 } \frac{\pi}{2} dθ \\ = \frac{\pi}{2} \int _{ - 4 }^{ - 1 } dθ$

We have a formula for integration as follows

$\int dt = t +c$

Using this we get

$I = \frac{\pi}{2} ( - 1 - ( - 4)) = \frac{3\pi}{2}$

Therefore,the value of given integral under given limits is

$\frac{3\pi}{2}$

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