What is the integral from 0 to x of e^{-2t^4} ?

The integral from 0 to x of e^{-2t^4} is

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integral from 0 to x of e^{-2t^4}

\int_{0}^{x} e^{- 2 t^{4}} d t

\frac{1}{4 2} - \frac{4 2 x Γ ( \frac{1}{4} , ( 4 2 x ) ^{4} )}{4 4 ( 4 2 x ) ^{4}} - (- \frac{0 \cdot Γ ( \frac{1}{4} , 0 ^{4} )}{4 4 0 ^{4}})

Solution steps

\int_{0}^{x} e^{- 2 t^{4}} d t

= \int_{0}^{x} e^{- (42 t)^{4}} d t

Apply u-substitution

: \int_{0}^{42 x} e^{- u^{4}} \frac{1}{4 2} d u

= \int_{0}^{42 x} e^{- u^{4}} \frac{1}{4 2} d u

Take the constant out:

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

= \frac{1}{4 2} \cdot \int_{0}^{42 x} e^{- u^{4}} d u

Use the nonelementary integral:

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

= \frac{1}{4 2} [- \frac{u Γ ( \frac{1}{4} , u ^{4} )}{4 4 u ^{4}}]_{0}^{42 x}

Compute the boundaries:

- \frac{4 2 x Γ ( \frac{1}{4} , ( 4 2 x ) ^{4} )}{4 4 ( 4 2 x ) ^{4}} - (- \frac{0 \cdot Γ ( \frac{1}{4} , 0 ^{4} )}{4 4 0 ^{4}})

= \frac{1}{4 2} - \frac{4 2 x Γ ( \frac{1}{4} , ( 4 2 x ) ^{4} )}{4 4 ( 4 2 x ) ^{4}} - (- \frac{0 \cdot Γ ( \frac{1}{4} , 0 ^{4} )}{4 4 0 ^{4}})

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What is the integral from 0 to x of e^{-2t^4} ?
The integral from 0 to x of e^{-2t^4} is