According to the Mean Value Theorem, for a continuous function f(x) in the interval [a, b], there exists a value $$\xi $$ in this interval such that $$\int\limits_{\text{a}}^{\text{b}} {{\text{f}}\left( {\text{x}} \right){\text{dx}} = } $$
A. $${\text{f}}\left( \xi \right)\left( {{\text{b}} - {\text{a}}} \right)$$
B. $${\text{f}}\left( {\text{b}} \right)\left( {\xi - {\text{a}}} \right)$$
C. $${\text{f}}\left( {\text{a}} \right)\left( {{\text{b}} - \xi } \right)$$
D. 0
Answer: Option A
The Taylor series expansion of 3 sinx + 2 cosx is . . . . . . . .
A. 2 + 3x - x2 - \[\frac{{{{\text{x}}^3}}}{2}\] + ...
B. 2 - 3x + x2 - \[\frac{{{{\text{x}}^3}}}{2}\] + ...
C. 2 + 3x + x2 + \[\frac{{{{\text{x}}^3}}}{2}\] + ...
D. 2 - 3x - x2 + \[\frac{{{{\text{x}}^3}}}{2}\] + ...
B. \[\infty \]
C. \[\frac{1}{2}\]
D. \[ - \infty \]
A. \[1 + \frac{{{{\left( {{\text{x}} - \pi } \right)}^2}}}{{3!}} + ...\]
B. \[ - 1 - \frac{{{{\left( {{\text{x}} - \pi } \right)}^2}}}{{3!}} + ...\]
C. \[1 - \frac{{{{\left( {{\text{x}} - \pi } \right)}^2}}}{{3!}} + ...\]
D. \[ - 1 + \frac{{{{\left( {{\text{x}} - \pi } \right)}^2}}}{{3!}} + ...\]

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