What is the integral from-1 to 1 of e^x-x^2+6 ?

The integral from-1 to 1 of e^x-x^2+6 is e+34/3-1/e

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integral from-1 to 1 of e^x-x^2+6

\int_{- 1}^{1} e^{x} - x^{2} + 6 d x

e + \frac{34}{3} - \frac{1}{e}

Decimal

13.68373 \dots

Solution steps

\int_{- 1}^{1} e^{x} - x^{2} + 6 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}^{1} e^{x} d x - \int_{- 1}^{1} x^{2} d x + \int_{- 1}^{1} 6 d x

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

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

\int_{- 1}^{1} 6 d x = 12

= e - \frac{1}{e} - \frac{2}{3} + 12

Combine the fractions

12 - \frac{2}{3} : \frac{34}{3}

= e + \frac{34}{3} - \frac{1}{e}

x \to 5 lim (\frac{2 - 25}{x - 5})

n = 1 \sum \infty \frac{1}{e ^{n + 1}}

(3, - 4), (5, 1)

\int \frac{x + 2}{x ^{2} - 1} d x

cos (2 θ)

What is the integral from-1 to 1 of e^x-x^2+6 ?
The integral from-1 to 1 of e^x-x^2+6 is e+34/3-1/e