Solve : ∫13 x2+x+2dx

Here, a=1, b=3nh=b−a=2f(x)=x2+x+2f(a+rh)=f(1+rh)(1+rh)2+(1+rh)+2=(rh)2+3(rh)+4Now, ∫13 x2+x+2dx=limh→0 h∑r=0n f(a+rh)=limh→0 h∑r=0n (rh)2+3(rh)+4=limh→0 hh2(n(n+1)(2n+1)6+3hn(n+1)2+4h=limh→0 nh(nh+h)(2nh+h)6+3nh(nh+h)2+4nh=limh→0 2(2+h(4+h)6+2(2+h)2+8=83+6+8=50/3

Solve : ∫13 x2+x+2dx

Here, a=1, b=3nh=b−a=2f(x)=x2+x+2f(a+rh)=f(1+rh)(1+rh)2+(1+rh)+2=(rh)2+3(rh)+4Now, ∫13 x2+x+2dx=limh→0 h∑r=0n f(a+rh)=limh→0 h∑r=0n (rh)2+3(rh)+4=limh→0 hh2(n(n+1)(2n+1)6+3hn(n+1)2+4h=limh→0 nh(nh+h)(2nh+h)6+3nh(nh+h)2+4nh=limh→0 2(2+h(4+h)6+2(2+h)2+8=83+6+8=50/3

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Solve : $\int_{1}^{3} (x^{2} + x + 2) d x$

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$- 50 / 3$

$50 / 3$

$\sqrt{50 / 3}$

$- \sqrt{50 / 3}$

answer is B.

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Detailed Solution

Here, $a = 1, b = 3$

$\begin{array}{l} n h = b - a = 2 \\ f (x) = (x^{2} + x + 2) \end{array}$

$f (a + r h) = f (1 + r h)$

$(1 + r h)^{2} + (1 + r h) + 2 = (r h)^{2} + 3 (r h) + 4$

Now, $\int_{1}^{3} (x^{2} + x + 2) d x$

$\begin{array}{l} = lim_{h \to 0} h \sum_{r = 0}^{n} f (a + r h) \\ = lim_{h \to 0} h \sum_{r = 0}^{n} ((r h)^{2} + 3 (r h) + 4) \\ = lim_{h \to 0} h (\frac{h^{2} (n (n + 1) (2 n + 1)}{6}) \\ + 3 h (\frac{n (n + 1)}{2 + 4 h}) \\ = lim_{h \to 0} (\frac{n h (n h + h) (2 n h + h)}{6} \\ (+ 3 (\frac{n h (n h + h)}{2}) + 4 n h) \\ = lim_{h \to 0} (\frac{2 (2 + h (4 + h)}{6} + (\frac{2 (2 + h)}{2}) + 8) \\ = \frac{8}{3} + 6 + 8 \\ = 50 / 3 \end{array}$