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

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

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

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

\frac{e ^{3} - e ^{\frac{3}{2}}}{3}

Decimal Notation

5.20128 \dots

Solution steps

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

Apply u-substitution

: \int_{3}^{\frac{3}{2}} - \frac{e ^{u}}{3} d u

= \int_{3}^{\frac{3}{2}} - \frac{e ^{u}}{3} d u

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

= - \int_{\frac{3}{2}}^{3} - \frac{e ^{u}}{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{3}{2}}^{3} e^{u} d u)

Use the common integral:

\int e^{u} d u = e^{u}

= - (- \frac{1}{3} [e^{u}]_{\frac{3}{2}}^{3})

Simplify

= \frac{1}{3} [e^{u}]_{\frac{3}{2}}^{3}

Compute the boundaries:

e^{3} - e^{\frac{3}{2}}

= \frac{1}{3} (e^{3} - e^{\frac{3}{2}})

Simplify

= \frac{e ^{3} - e ^{\frac{3}{2}}}{3}

y^{^{''}} + 16 y^{^{'}} + 64 y = 0

\int_{- 2}^{2} x^{3} + 1 d x

f (x) = x^{3} + x, at x = 0

cos^{2} (a x)

f^{^{'}} (s) = 10 s - 12 s^{3}, f (3) = 2

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