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

CBSE Class 12 · 2015 · Definite IntegralsLong Answer

Evaluate

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

as a limit of a sum.

Answer: $\frac{32}{3} + \frac{e ^{- 1} - e ^{- 7}}{3}$

Split into an exponential part and a polynomial part; each is a limit of a sum, and we give their values.
Polynomial part:

\int_{1}^{3} (x^{2} + 1) d x = [\frac{x ^{3}}{3} + x]_{1}^{3} = (9 + 3) - (\frac{1}{3} + 1) = \frac{32}{3}

.
Exponential part:

\int_{1}^{3} e^{2 - 3 x} d x = [\frac{e ^{2 - 3 x}}{- 3}]_{1}^{3} = - \frac{1}{3} (e^{- 7} - e^{- 1}) = \frac{e ^{- 1} - e ^{- 7}}{3}

I = \frac{32}{3} + \frac{e ^{- 1} - e ^{- 7}}{3}

CBSE Class 12 Board · 2015 · Subjective