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If \( f_{1}, f_{2}, \ldots f_{n} \) are integrable functions then \( \int_{a}^{b} \sum_{k=1}^{n} f_{k}(x) d x= \) \( \sum_{k=1}^{n} \int_{a}^{b} f_{k}(x) d x \). In certain cases, we can also obtain
\[
\begin{aligned}
\int_{a}^{b}\left(\sum_{k=0}^{\infty} a_{k} x^{k}\right) d x & =\sum_{k=0}^{\infty} \int_{a}^{b} a_{k} x^{k} d x \\
& =\sum_{k=0}^{\infty} \frac{a_{k}}{k+1}\left[b^{k+1}-a^{k+1}\right]
\end{aligned}
\]
e.g. we know that \( \frac{1}{1+x}=1-x+x^{2}-x^{3}+\ldots \) for \( x \in \)
\( (-1,1) \). Integrating both sides we get \( \log (1+x)=x \) \( -\ldots \). This term wise term integration can also be used in estimating the integral if the antiderivative is not known.
Approximation of \( \int_{0}^{1 / 2} \frac{1}{\sqrt{1+x^{2}}} d x \) within .001 is
(a) \( 0.4621 \)
(b) \( 0.479 \)
(c) \( 0.4234 \)
(d) \( 0.442 \)
\( \mathrm{P} \)
W
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