What is the integral from 0 to x of e^{x-t}y ?

The integral from 0 to x of e^{x-t}y is -y(1-e^x)

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integral from 0 to x of e^{x-t}y

\int_{0}^{x} e^{x - t} y d t

- y (1 - e^{x})

Solution steps

\int_{0}^{x} e^{x - t} y d t

Take the constant out:

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

= y \cdot \int_{0}^{x} e^{x - t} d t

Apply u-substitution

: \int_{x}^{0} - e^{u} d u

= y \cdot \int_{x}^{0} - e^{u} d u

Take the constant out:

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

= y (- \int_{x}^{0} e^{u} d u)

Use the common integral:

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

= y (- [e^{u}]_{x}^{0})

Simplify

= - y [e^{u}]_{x}^{0}

Compute the boundaries:

1 - e^{x}

= - y (1 - e^{x})

\int_{0}^{t} x e^{x} d x

\int_{0}^{1} 1 + t^{2} d t

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

\int_{2 x}^{x^{2}} e^{- t^{2}} d t

\int_{- 2}^{4} 6 + 4 x - x^{2} d x

What is the integral from 0 to x of e^{x-t}y ?
The integral from 0 to x of e^{x-t}y is -y(1-e^x)