What is the integral of (x^4)^{1/2} ?

The integral of (x^4)^{1/2} is (x^3)/3+C

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integral of (x^4)^{1/2}

\int (x^{4})^{\frac{1}{2}} d x

\frac{x ^{3}}{3} + C

Solution steps

\int (x^{4})^{\frac{1}{2}} d x

(x^{4})^{\frac{1}{2}} = x^{2},

assuming

x \geq 0

= \int x^{2} d x

Apply the Power Rule

: \frac{x ^{3}}{3}

= \frac{x ^{3}}{3}

Add a constant to the solution

= \frac{x ^{3}}{3} + C

h \to \frac{π}{2} lim (\frac{( 1 - cos ( 3 h ) )}{h})

\int x csc (x^{2}) d x

y^{^{''}} - 3 y^{^{'}} + 2 y = 2 e^{3 t}, y (0) = 0, y^{^{'}} (0) = 4

\frac{d}{d x} (x - ln (e^{x} cos (x)))