Question

INTEGRALS?!

Answer 1

\[\int\limits_{2}^{4}e^{3x}dx\]

Answer 2

\[\int\limits_{-2}^{2}\frac{ 1 }{ x^{2} }dx\]

Answer 3

\[\int_2^4e^{3x}~dx~~\Rightarrow~~\frac{1}{3}\int_6^{12} e^u~du=\frac{1}{3}\left(e^{12}-e^6\right)\]
\[\begin{align*}\int_{-2}^2\frac{dx}{x^2}&=\lim_{b\to0^-}\int_{-2}^b\frac{dx}{x^2}+\lim_{a\to0^+}\int_a^2\frac{dx}{x^2}\\&=\lim_{b\to0^-}\left[-\frac{1}{x}\right]_{-2}^b+\lim_{a\to0^+}\left[-\frac{1}{x}\right]_a^2\\
&=-\left(\lim_{b\to0^-}\frac{1}{b}-\frac{1}{2}\right)-\left(\frac{1}{2}-\lim_{a\to0^+}\frac{1}{a}\right)\\
&=-\lim_{b\to0^-}\frac{1}{b}+\lim_{a\to0^+}\frac{1}{a}
\end{align*}\]
Neither limit converges, so the integral diverges.