What is the integral from 0 to 1 of 4e^{7sqrt(x)} ?

The integral from 0 to 1 of 4e^{7sqrt(x)} is (8(6e^7+1))/(49)

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integral from 0 to 1 of 4e^{7sqrt(x)}

\int_{0}^{1} 4 e^{7 x} d x

\frac{8 ( 6 e ^{7} + 1 )}{49}

Decimal Notation

1074.41615 \dots

Solution steps

\int_{0}^{1} 4 e^{7 x} d x

Take the constant out:

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

= 4 \cdot \int_{0}^{1} e^{7 x} d x

Apply u-substitution

: \int_{0}^{7} \frac{2 e ^{u} u}{49} d u

= 4 \cdot \int_{0}^{7} \frac{2 e ^{u} u}{49} d u

Take the constant out:

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

= 4 \cdot \frac{2}{49} \cdot \int_{0}^{7} e^{u} u d u

Apply Integration By Parts

: [e^{u} u - \int e^{u} d u]_{0}^{7}

= 4 \cdot \frac{2}{49} [e^{u} u - \int e^{u} d u]_{0}^{7}

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

= 4 \cdot \frac{2}{49} [e^{u} u - e^{u}]_{0}^{7}

Simplify

4 \cdot \frac{2}{49} [e^{u} u - e^{u}]_{0}^{7} : \frac{8}{49} [e^{u} u - e^{u}]_{0}^{7}

= \frac{8}{49} [e^{u} u - e^{u}]_{0}^{7}

Compute the boundaries:

6 e^{7} + 1

= \frac{8}{49} (6 e^{7} + 1)

Simplify

= \frac{8 ( 6 e ^{7} + 1 )}{49}

1 - 6 x^{2} y^{^{'}} = x

x = - 2, y = x + 2, y = x^{2}

\int 8 y d y

m = 2 \sum \infty \frac{5}{6 ^{m}}

y^{^{'}} = 8 x^{6} y - y, y (1) = - 4

What is the integral from 0 to 1 of 4e^{7sqrt(x)} ?
The integral from 0 to 1 of 4e^{7sqrt(x)} is (8(6e^7+1))/(49)