Question

Find the laplace transform of L\left\{f(x)\right\} being f(x)=\int_0^3 \cos (3t)dt

Laplace transform
ANSWERED
asked 2021-09-10

Find the laplace transform of \(L\left\{f(x)\right\}\) being \(\displaystyle{f{{\left({x}\right)}}}={\int_{{0}}^{{3}}}{\cos{{\left({3}{t}\right)}}}{\left.{d}{t}\right.}\)

Expert Answers (1)

2021-09-11

Find \(L\left\{f(x)\right\}\) where \(\displaystyle{f{{\left({x}\right)}}}={\int_{{0}}^{{3}}}{\cos{{\left({3}{t}\right)}}}{\left.{d}{t}\right.}\)
Given that \(f(x)=\int_0^3 \cos(3t)dt\) , so we first calculate this integral.
\(\displaystyle{f{{\left({x}\right)}}}={\int_{{0}}^{{3}}}{\cos{{\left({3}{t}\right)}}}{\left.{d}{t}\right.}={{\left[{\frac{{{\sin{{\left({3}{t}\right)}}}}}{{{3}}}}\right]}_{{0}}^{{3}}}={\frac{{{1}}}{{{3}}}}{\left[{\sin{{9}}}-{\sin{{0}}}\right]}\)
\(\displaystyle\Rightarrow{f{{\left({x}\right)}}}={\frac{{{\sin{{9}}}}}{{{3}}}}\)
\(\displaystyle\Rightarrow{f{{\left({x}\right)}}}\) is a constant function
Now \(F(s)=L\left\{f(x)\right\}=\int_0^{\infty}f(x)e^{-sx}dx\)
\(\displaystyle={\int_{{0}}^{{\infty}}}{\frac{{{\sin{{9}}}}}{{{3}}}}{e}^{{-{s}{x}}}{\left.{d}{x}\right.}={\frac{{{\sin{{9}}}}}{{{3}}}}{\int_{{0}}^{{\infty}}}{e}^{{-{s}{x}}}{\left.{d}{x}\right.}\)
\(\displaystyle={\frac{{{\sin{{9}}}}}{{{3}}}}{{\left[{\frac{{{e}^{{-{s}{x}}}}}{{-{s}}}}\right]}_{{0}}^{{\infty}}}={\frac{{-{\sin{{9}}}}}{{{3}{s}}}}{\left[{e}^{{-\infty}}-{e}^{{0}}\right]}\)
\(\Rightarrow F(s)=L\left\{f(x)\right\}=\frac{1}{3s}\sin 9\)

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