Exercise · ZW-SCALAR-LINE-INTEGRAL-002 · ★☆☆ Basic

Compute \(\int_C x\,ds\), with \(\mathbf r(t)=\langle t,2t\rangle\), \(0\le t\le1\).

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  • B. \(\tfrac{\sqrt5}{2}\)

Explanation

\(\|\mathbf r'\|=\sqrt5\) and \(x=t\), so \(\int_0^1 t\sqrt5\,dt=\tfrac{\sqrt5}{2}\). Forgetting the factor \(\|\mathbf r'\|\) gives \(\tfrac12\).

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Review the concept: Line integral of a scalar function

The line integral of a scalar function \(f\) over a curve \(C\) parametrized by \(\mathbf r(t)\), \(a\le t\le b\), accumulates the values of \(f\) weighted by the arc length \(ds\).

\[\int_C f\,ds=\int_a^b f(\mathbf r(t))\,\|\mathbf r'(t)\|\,dt\]

More practice problems: line integral of a scalar function

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