What is the integral from-1 to 2 of 4y^3x ?

The integral from-1 to 2 of 4y^3x is 15x

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integral from-1 to 2 of 4y^3x

\int_{- 1}^{2} 4 y^{3} x d y

15 x

Solution steps

\int_{- 1}^{2} 4 y^{3} x d y

Take the constant out:

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

= 4 x \cdot \int_{- 1}^{2} y^{3} d y

Apply the Power Rule

: [\frac{y ^{4}}{4}]_{- 1}^{2}

= 4 x [\frac{y ^{4}}{4}]_{- 1}^{2}

Compute the boundaries:

\frac{15}{4}

= 4 x \frac{15}{4}

Simplify

= 15 x

\int 2 sin (x) cos (cos (x)) d x

y^{^{''}} - 5 y^{^{'}} + 4 y = 0, y (0) = 2, y^{^{'}} (0) = - 3

x \to n - lim (f (x))

\int_{0}^{π} 5 sin (x) d x

x \to 1 lim (\frac{x ^{2} - 4}{x - 2})

What is the integral from-1 to 2 of 4y^3x ?
The integral from-1 to 2 of 4y^3x is 15x