What is the integral from-3 to 1 of g(x) ?

The integral from-3 to 1 of g(x) is -4g

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integral from-3 to 1 of g(x)

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

- 4 g

Solution steps

\int_{- 3}^{1} gx d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= g \cdot \int_{- 3}^{1} x d x

Apply the Power Rule

: [\frac{x ^{2}}{2}]_{- 3}^{1}

= g [\frac{x ^{2}}{2}]_{- 3}^{1}

Compute the boundaries:

- 4

= g (- 4)

Simplify

= - 4 g

\int_{0}^{1} (- x^{2} - 3 x) d x

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\int_{0}^{y^{2}} y^{3} + 1 d x

\int_{- 2}^{- 1} 6 x (2 x^{2} - 6)^{3} d x

What is the integral from-3 to 1 of g(x) ?
The integral from-3 to 1 of g(x) is -4g