1.∫03(2+x2) dx=\int_0^3 (2+x^2)\,dx =∫03(2+x2)dx= expressed as a limit of suma.limn→∞[2n+1n+12+22+⋯+6n2n3]\lim_{n\to\infty}[\frac{2n+1}{n}+\frac{1^2+2^2+\cdots+6n^2}{n^3}]limn→∞[n2n+1+n312+22+⋯+6n2]b.limn→∞[3n+12+22+⋯+6n2n3]\lim_{n\to\infty}[\frac{3}{n}+\frac{1^2+2^2+\cdots+6n^2}{n^3}]limn→∞[n3+n312+22+⋯+6n2]c.limn→∞[6n+12+22+⋯+9n2n3]\lim_{n\to\infty}[\frac{6}{n}+\frac{1^2+2^2+\cdots+9n^2}{n^3}]limn→∞[n6+n312+22+⋯+9n2]d.limn→∞[3n+12+22+⋯+3n2n3]\lim_{n\to\infty}[\frac{3}{n}+\frac{1^2+2^2+\cdots+3n^2}{n^3}]limn→∞[n3+n312+22+⋯+3n2]Login to continueOnly logged in users canattempt or see the solution.
