Solution:- \(L\)\(({t^{2}e^{2t}+2\sin(t)\cos(t)+3})\) \(=\)\(L(t^{2}e^{2t})\) +\(L\)\((2\sin(t)\cos(t))\) +\(L(3)\) ——————————-(\(1)\) \(L(t^{2}e^{2t})\)\(=L(t^{2})\).\(L(e^{2t})\) \(=\frac{2}{s^3}.\frac{1}{S-2}\), ——————————————————–(\(a\)) Since, \(L({t^n})\) \(=\) \(\frac{n!}{S^{n+1}}\) \(\Rightarrow\)\(L(t^{2})\)\(=\) \(\frac{2}{S^{2+1}}\) \(=\) \(\frac{2}{S^{3}}\) Since, \(L(e^{at})\)\(=\)\(\frac{1}{(S-a)}\)\(=\)\(\frac{1}{S-2}\) Now, \(L({2\sin(t)\cos(t)})\) Which can be written as \(=L({\sin2(t)})\) \(L(\sin2t)\)\(=\)\(\frac{2}{S^{2}+2^{2}}\) \(=\)\(\frac{2}{(S^{2}+4)}\) ———————————————————–(\(b\)) Since, Laplace transformation of \(L\)\((\sin\)\(at\)) \(=\)\(\frac{a}{S^{2}+a^{2}}\) now, \(L(3)\)\(=\frac{3}{S}\) ———————————————————-(\(c\)) Since, \(L(1)\)\(=\frac{1}{S}\) Now putting the values of (\(a)\), Now putting the values of (\(a)\), Now putting […]

Solution:- (a) (frac{x^{2}}{4} +frac{y^{2}}{9}=1) (a=2) and (b=3) from the above relation by changing the variable taking, (u= rcosTheta) and (v=rsinTheta) Jacobian ((J))(=) (=)(frac{du}{dr}) (frac{du}{dtheta}) (frac{dv}{dr}) (frac{dv}{dtheta}) (=)(cos) (-rsintheta) (sintheta) (rcostheta) (=) (rcos^{2}theta+rsin^{2}theta) (=)(r)((sin^{2}theta+cos^{2}theta)) (J)(=)(r).(1) (J)(=)(r) Limits are, (r=0) to(1) and 0 to (2pi) since radius is (1) Area (=) (ab) (int_{0}^{2pi}int_{0}^{1})(rdrdtheta) (=)(2.3int_{0}^{2pi}[r^{2}]_{0}^{1}) (=)(6int_{0}^{2pi}(1-0)dtheta) (=)(6.[theta]_{0}^{2pi}) (=)(frac{12Pi}{2}) […]

Solution:- \(x^{3}+y^{3}-1=0\) \(x^{3}+y^{3}=1\) now, \(\frac{d}{dx}(x^{3}+y^{3})=\frac{d}{dx}(1)\) \(\frac{d}{dx}(x^{3})\)+ \(\frac{d}{dx}(y^{3})= 0\) \(3x^{2}+3y^{2} \frac{d}{dx}=0\) \(3y^{2}\frac{d}{dx}= – 3x^{2}\) \(\frac{d}{dx}\)\(=\)\(-\)\(\frac{x^{2}}{y^{2}}\)

Solution: To solve this question we can take various approach but here we are taking the unit price approach and compare their price to determine which is the better buy (17) ounce can costs (=) ($) (2.49) therefor, (1) ounce can costs (=)(frac{2.49}{17}) (=) ($)(0.146) per ounce (26) ounce can costs (=) ($2.89) therefore, (1) […]

(intfrac{x+cosx}{3x^{2}+6sinx})(dx) Let, (3x^{2}+6sinx) (= t) Differentiating w.r.t (x) (6x+6cosx) (dx) (=dt) (6(x+cosx)) (dx) (=dt) ((x+cosx))(dx)(=)(intfrac{1}{6})(dt) (intfrac{x+cosx}{3x^{2}+6sinx})(=)(intfrac{1}{6})(frac{dt}{t}) (=)(frac{1}{6})(intfrac{dt}{t}) (=)(frac{1}{6})(log|t|+C) Putting the value of (t) = (3x^{2}+6sinx) (=)(frac{1}{6})(log|3x^{2}+6sinx|+C)