What is the integral from a to b of [3f(t)-2g(t)] ?

The integral from a to b of [3f(t)-2g(t)] is -g(b^2-a^2)+(3f(b^2-a^2))/2

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integral from a to b of [3f(t)-2g(t)]

\int_{a}^{b} [3 f (t) - 2 g (t)] d t

- g (b^{2} - a^{2}) + \frac{3 f ( b ^{2} - a ^{2} )}{2}

Solution steps

\int_{a}^{b} 3 f t - 2 g t d t

Apply the Sum Rule:

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

= - \int_{a}^{b} 2 g t d t + \int_{a}^{b} 3 f t d t

\int_{a}^{b} 2 g t d t = g (b^{2} - a^{2})

\int_{a}^{b} 3 f t d t = \frac{3 f ( b ^{2} - a ^{2} )}{2}

= - g (b^{2} - a^{2}) + \frac{3 f ( b ^{2} - a ^{2} )}{2}

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What is the integral from a to b of [3f(t)-2g(t)] ?
The integral from a to b of [3f(t)-2g(t)] is -g(b^2-a^2)+(3f(b^2-a^2))/2