What is the integral from 1 to 2 of (3-e^x) ?

The integral from 1 to 2 of (3-e^x) is 3-e^2+e

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integral from 1 to 2 of (3-e^x)

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

3 - e^{2} + e

Decimal

- 1.67077 \dots

Solution steps

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{1}^{2} 3 d x - \int_{1}^{2} e^{x} d x

\int_{1}^{2} 3 d x = 3

\int_{1}^{2} e^{x} d x = e^{2} - e

= 3 - (e^{2} - e)

Simplify

= 3 - e^{2} + e

y^{^{''}} + 4 y = e^{t} (cos (2 t) + sin (2 t))

f (x) = - 7 x^{2}

x \to 0 lim ((ln (\frac{1}{x}))^{x})

\int (2 x + 3)^{7} d x

n = 0 \sum \infty - 2.04258 E 15

What is the integral from 1 to 2 of (3-e^x) ?
The integral from 1 to 2 of (3-e^x) is 3-e^2+e