What is the integral from 1 to 7 of e^{x^{-3}} ?

The integral from 1 to 7 of e^{x^{-3}} is

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integral from 1 to 7 of e^{x^{-3}}

\int_{1}^{7} e^{x^{- 3}} d x

Solution steps

\int_{1}^{7} e^{x^{- 3}} d x

Apply u-substitution

: \int_{1}^{\frac{1}{343}} - \frac{e ^{u}}{3 u ^{\frac{4}{3}}} d u

= \int_{1}^{\frac{1}{343}} - \frac{e ^{u}}{3 u ^{\frac{4}{3}}} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{\frac{1}{343}}^{1} - \frac{e ^{u}}{3 u ^{\frac{4}{3}}} d u

Take the constant out:

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

= - (- \frac{1}{3} \cdot \int_{\frac{1}{343}}^{1} \frac{e ^{u}}{u ^{\frac{4}{3}}} d u)

Apply Integration By Parts

Compute the boundaries:

\int_{2}^{3} cos^{5} (y) d y

\int_{0}^{1000} e^{4 x} - 5 d x

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

\int_{2}^{5} (t^{2} + 2 t) d t

\int_{0}^{2} y e^{- y^{2}} d y

What is the integral from 1 to 7 of e^{x^{-3}} ?
The integral from 1 to 7 of e^{x^{-3}} is