Evaluate: $\int_{1}^{3}\left(x^{2}+3 x+e^{x}\right) d x$, as the limit of the sum.

CBSE Class 12 · 2018 · Definite IntegralsLong Answer

Evaluate:

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

, as the limit of the sum.

Answer: $\frac{62}{3} + e^{3} - e$

With

a = 1, b = 3, h = \frac{2}{n}, nh = 2

and

f (x) = x^{2} + 3 x + e^{x}

, use

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

.
Polynomial part:

\int_{1}^{3} (x^{2} + 3 x) d x = [\frac{x ^{3}}{3} + \frac{3 x ^{2}}{2}]_{1}^{3} = \frac{45}{2} - \frac{11}{6} = \frac{62}{3}

.
Exponential part:

\sum_{r = 0}^{n - 1} e^{1 + r h} = e \cdot \frac{e ^{nh} - 1}{e ^{h} - 1}

, and

h \cdot \frac{1}{e ^{h} - 1} \to 1

h \to 0

, so

h \sum \to e (e^{2} - 1) = e^{3} - e

I = \frac{62}{3} + e^{3} - e

CBSE Class 12 Board · 2018 · Subjective