What is the integral from 0 to 6 of e^{x/2} ?

The integral from 0 to 6 of e^{x/2} is 2(e^3-1)

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integral from 0 to 6 of e^{x/2}

\int_{0}^{6} e^{\frac{x}{2}} d x

2 (e^{3} - 1)

Decimal

38.17107 \dots

Solution steps

\int_{0}^{6} e^{\frac{x}{2}} d x

Apply u-substitution

: \int_{0}^{3} e^{u} \cdot 2 d u

= \int_{0}^{3} e^{u} \cdot 2 d u

Take the constant out:

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

= 2 \cdot \int_{0}^{3} e^{u} d u

Use the common integral:

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

= 2 [e^{u}]_{0}^{3}

Compute the boundaries:

e^{3} - 1

= 2 (e^{3} - 1)

\int_{- \frac{π}{2}}^{π} 7 cos (x) d x

\int_{0}^{1} x \cdot sin (nπ x) d x

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

\int_{0}^{2} x^{3} - x^{2} - 2 x d x

\int_{0}^{1} y d y

What is the integral from 0 to 6 of e^{x/2} ?
The integral from 0 to 6 of e^{x/2} is 2(e^3-1)