What is the integral from-3 to 0 of sqrt(1-x)+1+x ?

The integral from-3 to 0 of sqrt(1-x)+1+x is 19/6

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integral from-3 to 0 of sqrt(1-x)+1+x

\int_{- 3}^{0} 1 - x + 1 + x d x

\frac{19}{6}

Decimal

3.16666 \dots

Solution steps

\int_{- 3}^{0} 1 - x + 1 + 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_{- 3}^{0} 1 - x d x + \int_{- 3}^{0} 1 d x + \int_{- 3}^{0} x d x

\int_{- 3}^{0} 1 - x d x = \frac{14}{3}

\int_{- 3}^{0} 1 d x = 3

\int_{- 3}^{0} x d x = - \frac{9}{2}

= \frac{14}{3} + 3 - \frac{9}{2}

Simplify

\frac{14}{3} + 3 - \frac{9}{2} : \frac{19}{6}

= \frac{19}{6}

\int_{3}^{\infty} \frac{1}{x ^{2} + 9} d x

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

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\int_{3}^{x^{3}} t^{2} d t

What is the integral from-3 to 0 of sqrt(1-x)+1+x ?
The integral from-3 to 0 of sqrt(1-x)+1+x is 19/6