What is the integral from y/2 to 1 of (ye^{x^3}) ?

The integral from y/2 to 1 of (ye^{x^3}) is

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integral from y/2 to 1 of (ye^{x^3})

\int_{\frac{y}{2}}^{1} (y e^{x^{3}}) d x

y - \frac{1 \cdot Γ ( \frac{1}{3} , - 1 ^{3} )}{3 3 - 1 ^{3}} - - \frac{\frac{y}{2} Γ ( \frac{1}{3} , - ( \frac{y}{2} ) ^{3} )}{3 3 - ( \frac{y}{2} ) ^{3}}

Solution steps

\int_{\frac{y}{2}}^{1} y e^{x^{3}} d x

Take the constant out:

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

= y \cdot \int_{\frac{y}{2}}^{1} e^{x^{3}} d x

Use the nonelementary integral:

\int e^{x^{a}} d x = - \frac{x Γ ( \frac{1}{a} , - x ^{a} )}{a a - x ^{a}}

= y [- \frac{x Γ ( \frac{1}{3} , - x ^{3} )}{3 3 - x ^{3}}]_{\frac{y}{2}}^{1}

Compute the boundaries:

- \frac{1 \cdot Γ ( \frac{1}{3} , - 1 ^{3} )}{3 3 - 1 ^{3}} - - \frac{\frac{y}{2} Γ ( \frac{1}{3} , - ( \frac{y}{2} ) ^{3} )}{3 3 - ( \frac{y}{2} ) ^{3}}

= y - \frac{1 \cdot Γ ( \frac{1}{3} , - 1 ^{3} )}{3 3 - 1 ^{3}} - - \frac{\frac{y}{2} Γ ( \frac{1}{3} , - ( \frac{y}{2} ) ^{3} )}{3 3 - ( \frac{y}{2} ) ^{3}}

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What is the integral from y/2 to 1 of (ye^{x^3}) ?
The integral from y/2 to 1 of (ye^{x^3}) is