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3A, , , , , , , , , , , , , , , UNIT-3, INTEGRAL CALCULUS, , , , , , , , PART-A, , , , , , , , , , , , , , , , ; PROBLEMS BASED ON DEFINITE INTEGRALS, , , , , , Define the term definite integral., , The definite integral’ of f from a to b is, , b : n, Sf f@)d = lim SY fog) Ax, , xe j=], Write down the mid ‘point rule in the definite integral., Midpoint Rule : ., , b : n Sf@yae =F fq Ax = Ax[fGr) +-- +f) |, , i, , b-a, n, , , , Ax =, , - 4, , and Xj = 5 Oi +x) = midpoint of [x1], , f calculus part-1., , 1. If f is continuous, , State the fundamental theorem 0, , The Fundemental Theorem of Calculus, Parton [a,b], then the function g is defined by, , g(x) = Ssoasasx xsb, , is continuous on {a,b} and differentiable on ( b), and, , 8°G)=f@)

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3, wreGRAL CALCULUS af, , ;, What is wrong with the equation ?, , f secOtandd@ = [seco] =3, , wh \a/3, , Solution : The function f (6) = sec @ tan @ is not: continuous on the, , interval ie “| so FTC2 cannot be applied. ['. tan> = 00], , 8. Prove that following integral by interpreting each in-terms of, areas, , , , [A.U N/D 2016, MATI51, R-15, U.D], , . b ., Solution : Given : f xdx; Here, f@ =x;, a, 2 x, Antiderivative f(x) = a, , 2 2 2 ge, b b-., Area = F(b)-F(@) = 5-5 = 4, , , , by FTC2.