What is the integral from-3 to 5 of 6-y-((y-3)^2)/4 ?

The integral from-3 to 5 of 6-y-((y-3)^2)/4 is 64/3

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integral from-3 to 5 of 6-y-((y-3)^2)/4

\int_{- 3}^{5} 6 - y - \frac{( y - 3 ) ^{2}}{4} d y

\frac{64}{3}

Solution steps

\int_{- 3}^{5} 6 - y - \frac{( y - 3 ) ^{2}}{4} d y

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}^{5} 6 d y - \int_{- 3}^{5} y d y - \int_{- 3}^{5} \frac{( y - 3 ) ^{2}}{4} d y

\int_{- 3}^{5} 6 d y = 48

\int_{- 3}^{5} y d y = 8

\int_{- 3}^{5} \frac{( y - 3 ) ^{2}}{4} d y = \frac{56}{3}

= 48 - 8 - \frac{56}{3}

Simplify

48 - 8 - \frac{56}{3} : \frac{64}{3}

= \frac{64}{3}

f (x) = 7 tan^{3} (x)

1 6^{x}

\int + \frac{16}{( 64 x ^{2} + 1 ) ^{2}} d x

\int_{- \infty}^{0} \frac{1}{4 - 5 x} d x

x \to - 1 lim (\frac{x + 3}{x ^{2} + 1})

What is the integral from-3 to 5 of 6-y-((y-3)^2)/4 ?
The integral from-3 to 5 of 6-y-((y-3)^2)/4 is 64/3