Let x be a continuous variable defined over the interval $$\left( { - \infty ,\,\infty } \right)$$ , and f(x) = e-x-e-x . The integral $${\text{g}}\left( {\text{x}} \right) = \int {{\text{f}}\left( {\text{x}} \right){\text{dx}}} $$ is equal to
A. ee-x
B. e-e-x
C. e-ex
D. e-x
Answer: Option B
Related Questions on Calculus
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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