What is the integral from 0.25 to 1.25 of e^{-4x^2} ?

The integral from 0.25 to 1.25 of e^{-4x^2} is 0.21229…

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integral from 0.25 to 1.25 of e^{-4x^2}

\int_{0.25}^{1.25} e^{- 4 x^{2}} d x

0.21229 \dots

Solution steps

\int_{0.25}^{1.25} e^{- 4 x^{2}} d x

= \int_{0.25}^{1.25} e^{- (4 x)^{2}} d x

Apply u-substitution

: \int_{0.5}^{2.5} e^{- u^{2}} \frac{1}{2} d u

= \int_{0.5}^{2.5} e^{- u^{2}} \frac{1}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int_{0.5}^{2.5} e^{- u^{2}} d u

Use the nonelementary integral:

\int e^{- u^{2}} d u = \frac{π}{2} erf (u)

= \frac{1}{2} [\frac{π}{2} erf (u)]_{0.5}^{2.5}

Simplify

= 0.5 [0.88622 \dots erf (u)]_{0.5}^{2.5}

Compute the boundaries:

0.42458 \dots

= 0.5 \cdot 0.42458 \dots

Simplify

= 0.21229 \dots

\int_{0}^{2} \frac{1}{2} e^{- \frac{x}{2}} d x

\int_{0}^{1} 5 x (x^{2} - 1)^{5} d x

\int_{1}^{2} (2 x + 6) d x

\int_{0}^{1} x^{3} e^{- x} d x

\int_{0}^{4} \frac{1}{36 x ^{2} + 49} d x

What is the integral from 0.25 to 1.25 of e^{-4x^2} ?
The integral from 0.25 to 1.25 of e^{-4x^2} is 0.21229…