What is the integral of-8(y)^2(1) ?

The integral of-8(y)^2(1) is -(8y^3)/3+C

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integral of-8(y)^2(1)

\int - 8 (y)^{2} (1) d y

- \frac{8 y ^{3}}{3} + C

Solution steps

\int - 8 y^{2} \cdot 1 d y

Take the constant out:

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

= - 8 \cdot 1 \cdot \int y^{2} d y

Simplify

= - 8 \cdot \int y^{2} d y

Apply the Power Rule

: \frac{y ^{3}}{3}

= - 8 \cdot \frac{y ^{3}}{3}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= - \frac{8 y ^{3}}{3}

Add a constant to the solution

= - \frac{8 y ^{3}}{3} + C

x = y^{2} + 14 y + 16, x = 7 y + 6

2 x + 1

\int + \frac{1}{( 9 x ^{2} + 1 ) ^{\frac{5}{2}}} d x

f (x) = \frac{6 x}{( x ^{3} - 9 ) ^{2}}

x \to 3 + lim (2 x + 5)