What is the integral from 4 to 2 of g(x) ?

The integral from 4 to 2 of g(x) is -g*6

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integral from 4 to 2 of g(x)

\int_{4}^{2} g (x) d x

- g \cdot 6

Solution steps

\int_{4}^{2} gx d x

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{2}^{4} gx d x

Take the constant out:

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

= - g \cdot \int_{2}^{4} x d x

Apply the Power Rule

: [\frac{x ^{2}}{2}]_{2}^{4}

= - g [\frac{x ^{2}}{2}]_{2}^{4}

Compute the boundaries:

6

= - g \cdot 6

\int_{\frac{π}{3}}^{x} cos^{2} (t) d t

\int_{0}^{2} (3 t - 1)^{4} d t

\int_{\frac{π}{4}}^{\frac{3 π}{4}} sin (x) d x

\int_{0}^{1} \frac{1}{1 - 4 x ^{2}} d x

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

What is the integral from 4 to 2 of g(x) ?
The integral from 4 to 2 of g(x) is -g*6