The anti derivative of sin 2x – 4e3x is the function of x whose derivative is sin 2x – 4e3x.
It is known that,
\begin{align} \frac {d}{dx} \left(-\frac{1}{2}cos 2x – \frac {4}{3} e^{3x}\right) = sin2x – 4e^{3x} \end{align}
Therefore, the anti derivative of (sin 2x – 4e3x) is \begin{align} \left(-\frac{1}{2}cos 2x – \frac {4}{3} e^{3x}\right) \end{align}
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Find the area of the region bounded by the curve y2 = x and the lines x = 1, x = 4 and the x-axis.
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y = ex +1 : yn -y' = 0
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(A) 10π (B) 12π (C) 8π (D) 11π
If a line has the direction ratios −18, 12, −4, then what are its direction cosines?
Let f : N → N be defined by
State whether the function f is bijective. Justify your answer.