What is the integral from-ln(4) to 0 of 6sqrt(3)e^t ?

The integral from-ln(4) to 0 of 6sqrt(3)e^t is (9sqrt(3))/2

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integral from-ln(4) to 0 of 6sqrt(3)e^t

\int_{- l n (4)}^{0} 63 e^{t} d t

\frac{9 3}{2}

Solution steps

\int_{- l n (4)}^{0} 63 e^{t} d t

Take the constant out:

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

= 63 \cdot \int_{- l n (4)}^{0} e^{t} d t

Use the common integral:

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

= 63 [e^{t}]_{- l n (4)}^{0}

Simplify

ln (4) : 2 ln (2)

= 63 [e^{t}]_{- 2 l n (2)}^{0}

Compute the boundaries:

\frac{3}{4}

= 63 \frac{3}{4}

Simplify

= \frac{9 3}{2}

\frac{d}{d x} (\frac{e ^{x}}{x + 2})

f (x) = (2 x^{2} - 4)^{4}

\int \frac{1}{x ^{2} + 2} d (x^{2} + 1) d x

x \to \infty lim (2 x!)

x \to 0 lim ((x - 1)^{x - 1})

What is the integral from-ln(4) to 0 of 6sqrt(3)e^t ?
The integral from-ln(4) to 0 of 6sqrt(3)e^t is (9sqrt(3))/2