Solve \int _(0)^(t)2e^(2t-4u)du

Question

Evaluate the integral

\frac{- 1 + e ^{4 t}}{2 e ^{2 t}}

Evaluate

\int_{0}^{t} 2 e^{2 t - 4 u} d u

Evaluate the integral

\int 2 e^{2 t - 4 u} d u

Use the property of integral \int k f (x) d x = k \int f (x) d x

2 \times \int e^{2 t - 4 u} d u

Use the substitution d u = - \frac{1}{4} d t to transform the integral

More Steps

2 \times \int e^{2 t - 4 u} (- \frac{1}{4}) d t

Simplify

More Steps

2 \times \int - \frac{e ^{2 t - 4 u}}{4} d t

Use the substitution t = 2 t - 4 u to transform the integral

2 \times \int \frac{- e ^{t}}{4} d t

Simplify the expression

2 \times \int - \frac{e ^{t}}{4} d t

Rewrite the expression

2 \times \int - \frac{1}{4} e^{t} d t

Use the property of integral \int k f (x) d x = k \int f (x) d x

2 (- \frac{1}{4}) \times \int e^{t} d t

Multiply the numbers

More Steps

- \frac{1}{2} \times \int e^{t} d t

Use the property of integral \int e^{x} d x = e^{x}

- \frac{1}{2} e^{t}

Multiply the terms

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

Substitute back

- \frac{e ^{2 t - 4 u}}{2}

Return the limits

(- \frac{e ^{2 t - 4 u}}{2})_{0}^{t}

Calculate the value

More Steps

- \frac{1}{2 e ^{2 t}} + \frac{e ^{2 t}}{2}

Reduce fractions to a common denominator

- \frac{1}{2 e ^{2 t}} + \frac{e ^{2 t} \times e ^{2 t}}{2 e ^{2 t}}

Write all numerators above the common denominator

\frac{- 1 + e ^{2 t} \times e ^{2 t}}{2 e ^{2 t}}

Solution

More Steps

\frac{- 1 + e ^{4 t}}{2 e ^{2 t}}

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