What is the integral from e^{3x-6}+10 to 11 of 1 ?

The integral from e^{3x-6}+10 to 11 of 1 is -e^{3x-6}+1

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integral from e^{3x-6}+10 to 11 of 1

\int_{e^{3 x - 6} + 10}^{11} 1 d y

- e^{3 x - 6} + 1

Solution steps

\int_{e^{3 x - 6} + 10}^{11} 1 d y

Integral of a constant:

\int a d x = a x

= [1 \cdot y]_{10 + e^{- 6 + 3 x}}^{11}

Simplify

= [y]_{10 + e^{- 6 + 3 x}}^{11}

Compute the boundaries:

- e^{3 x - 6} + 1

= - e^{3 x - 6} + 1

\int_{1}^{e^{1}} \frac{1}{x} d x

\int_{0}^{1} \frac{2}{x + 1} d x

\int_{0}^{y - z} 2 x - y d x

\int_{- 4}^{4} (4 + 16 - x^{2}) d x

\int_{1}^{x^{2}} e^{t^{2}} d t

What is the integral from e^{3x-6}+10 to 11 of 1 ?
The integral from e^{3x-6}+10 to 11 of 1 is -e^{3x-6}+1