Evaluate $\int_{1}^{3}\left(2 x^{2}+5\right) d x$ as a limit of a sum

CBSE Class 12 · 2012 · Definite IntegralsLong Answer

Evaluate

\int_{1}^{3} (2 x^{2} + 5) d x

as a limit of a sum

Answer: $\frac{82}{3}$

Here

f (x) = 2 x^{2} + 5

a = 1

b = 3

, so

h = \frac{b - a}{n} = \frac{2}{n}

and

nh = 2

.
Use

\int_{a}^{b} f (x) d x = lim_{h \to 0} h \sum_{r = 0}^{n - 1} f (a + r h)

, with

f (1 + r h) = 2 (1 + r h)^{2} + 5 = 7 + 4 r h + 2 r^{2} h^{2}

r = 0 \sum n - 1 f (1 + r h) = 7 n + 4 h \cdot \frac{( n - 1 ) n}{2} + 2 h^{2} \cdot \frac{( n - 1 ) n ( 2 n - 1 )}{6}

Multiply by

h

and write everything through

nh = 2

h \sum = 7 (nh) + 2 (nh)^{2} (1 - \frac{1}{n}) + \frac{1}{3} (nh)^{3} (1 - \frac{1}{n}) (2 - \frac{1}{n})

Letting

n \to \infty

(so

h \to 0

I = 7 (2) + 2 (4) + \frac{1}{3} (8) (1) (2) = 14 + 8 + \frac{16}{3} = \frac{82}{3}

CBSE Class 12 Board · 2012 · Subjective