Answer Happy

Posted: **Thu Jul 14, 2022 4:04 pm**

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(तה) (a) Find g(t) so that ∫g(t)dt=2t4−t2+2⋅sin−12t 4−t2 −π 4+t2 π 2π 32π−23 32π+23 (b) Evaluate ∫02g(t) dt using the Evaluation Theorem. π 4+t2 32π−23 32π+23 4−t2 −π 2π
(c) (c) Evaluate ∫02g(t) dt by the area intepretation of the integral. Compare your answer with part (b) above 32π−23 π −π 4+t2 4+t2 32π+23 2π (d) (d) Evaluate ∫038(t)dt. 2π −π 32π−23 π 4−t2 4+t2 32π+23

Answer Happy

(तה) (a) Find g(t) so that ∫g(t)dt=2t​4−t2​+2⋅sin−12t​ 4−t2​ −π 4+t2​ π 2π 32π​−23​​ 32π​+23​​ (b) Evaluate ∫02​g(t) dt

(तה) (a) Find g(t) so that ∫g(t)dt=2t​4−t2​+2⋅sin−12t​ 4−t2​ −π 4+t2​ π 2π 32π​−23​​ 32π​+23​​ (b) Evaluate ∫02​g(t) dt

(तה) (a) Find g(t) so that ∫g(t)dt=2t4−t2+2⋅sin−12t 4−t2 −π 4+t2 π 2π 32π−23 32π+23 (b) Evaluate ∫02g(t) dt

(तה) (a) Find g(t) so that ∫g(t)dt=2t4−t2+2⋅sin−12t 4−t2 −π 4+t2 π 2π 32π−23 32π+23 (b) Evaluate ∫02g(t) dt