Integral Calculus (Integration) - SS3 Mathematics Past Questions and Answers - page 1

Let \(u = x^{2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \therefore\frac{du}{dx} = 2x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \therefore dx = \frac{du}{2x}\)

Substituting we have,

\(\int_{}^{}\frac{1}{x^{2}}\ dx = \ \int_{}^{}\frac{1}{u}.\frac{du}{2x}\ = \frac{1}{2x}\left( \ln u \right) + C\), substitute \(u\)

\[= \frac{1}{2x}\ln x^{2} + C\]

\(\therefore\int_{}^{}\frac{1}{x^{2}}dx = \frac{\ln x^{2}}{2x} + C\ \)

Let \(u = x^{2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \therefore\frac{du}{dx} = 2x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \therefore dx = \frac{du}{2x}\)

Substituting we have,

\(\int_{}^{}\frac{1}{x^{2}}\ dx = \ \int_{}^{}\frac{1}{u}.\frac{du}{2x}\ = \frac{1}{2x}\left( \ln u \right) + C\), substitute \(u\)

\[= \frac{1}{2x}\ln x^{2} + C\]

\(\therefore\int_{}^{}\frac{1}{x^{2}}dx = \frac{\ln x^{2}}{2x} + C\ \)

\[\int_{}^{}u\ dv = uv - \ \int_{}^{}v\ du\]

\(u = (x - 3)\), \(\frac{du}{dx} = 1\)

\(\frac{dv}{dx} = (x + 3)\), \(v = \frac{x^{2}}{2} + 3x\)

\(\int_{}^{}{(x - 3)(x + 3)}dx = (x - 3)\left( \frac{x^{2}}{2} + 3x \right) - \int_{}^{}\left( \frac{x^{2}}{2} + 3x \right)(1)\ \)

\[= (x - 3)\left( x^{2} + 6x \right) - \int_{}^{}{x^{2} + 6x}\]

\[= x^{3} + 3x^{2} - 18x - \frac{x^{3}}{3} + \frac{6x^{2}}{2}\]

\[= x^{3} + 3x^{2} - 18x - \frac{x^{3}}{3} + 3x^{2}\]

\[= x^{3} + 6x^{2} - 18x - \frac{x^{3}}{3}\]

\[= 3x^{3} + 18x^{2} - 54x - x^{3}\]

\[= 2x^{3} + 18x^{2} - 54x\]

\(= 2x\lbrack x^{2} + 9x - 27\rbrack\)

\(\int_{1}^{2}{2x^{3}}dx = \ \left\lbrack \frac{2x^{4}}{4} \right\rbrack_{1}^{2} = \left\lbrack \frac{x^{4}}{2} \right\rbrack_{1}^{2} = \left\lbrack \frac{{(2)}^{4}}{2} \right\rbrack - \left\lbrack \frac{(1)^{4}}{2} \right\rbrack = \frac{16}{2} - \frac{1}{2} = \frac{15}{2} = 7.5\)

First integrate \(\int_{}^{}e^{\frac{x}{2}}dx\), let \(u = \frac{x}{2}\), \(\frac{du}{dx} = \frac{2(1) - x(0)}{4} = \frac{2}{4} = \frac{1}{2}\), \(dx = 4\ du\)

\[\int_{}^{}e^{u}4\ du = 4\int_{}^{}e^{u}du = 4\ e^{u} = 4e^{\frac{x}{2}}\ \]

\[\int_{0}^{2}e^{\frac{x}{2}}dx = \ \left\lbrack 4e^{\frac{x}{2}} \right\rbrack_{0}^{2} = \left\lbrack 4e^{\frac{2}{2}} \right\rbrack - \left\lbrack 4e^{\frac{0}{2}} \right\rbrack = 4e - 4(1) = 4e - 4 = 4(e - 1)\]

\[\int_{}^{}u\ dv = uv - \ \int_{}^{}v\ du\]

\(u = x\), \(du = 1\)

\(dv = \sin x\), \(v = - \cos x\)

\[\int_{}^{}x\sin x\ dx = - x\cos x - \ \int_{}^{}{- \cos x}\]

\[= - x\cos x + \int_{}^{}{\cos x}\]

\[= - x\cos x + \sin x\]

\[= \sin x - x\cos x\]