Physics, asked by daman2550, 9 months ago

Integrate the following function (a) int_(o)^(2) 2t dt (b) int _(pi//6)^(pi//3) sin x dx (c) int _(4)^(10)(dx)/(x) (d) int _(o)^(pi) cos x dx (e) int _(1) ^(2)(2t -4) dt

Answers

Answered by NirmalPandya
1

(a) \int\limits^2_0 {2t} \, dt = 2\int\limits^2_0 {t} \, dt  

              = 2.[t²/2]₀²

              = [2²-0²]

              = 4

(b) \int\limits^\frac{\pi }{3} _\frac{\pi }{6}  {sinx} \, dx = [-cosx]\left \{ {{\frac{\pi }{3} } \atop {\frac{\pi }{6} }} \right.

                    = [-cos(π/3) - (-cos(π/6))]

                    = [-(1/2) + (\sqrt{3}/2)]

                    = \frac{\sqrt{3}-1 }{2}

(c) \int\limits^{10}_4 {\frac{dx}{x} } \, = [㏑x]₄¹⁰

           = ㏑10 - ㏑4

           = ㏑\frac{10}{4}

           = ㏑2.5

(d) \int\limits^\pi _0 {cosx} \, dx = [sinx]\left \{ {{\pi } \atop {0}} \right.

                   = sinπ - sin0

                   = 0

(e) \int\limits^2_1 {2t-4} \, dt = \int\limits^2_1 {2t} \, dt - \int\limits^2_1 {4} \, dt

                     = 2 [ t²/2 ]₁² - 4[t]₁²

                     = [ 2²- 1² ] - 4 [ 2 - 1 ]

                     = 3 - 4

                     = -1

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