Evaluate the definite integral. ∫27e2−xdx

Question

Evaluate the definite integral.  ∫27e2−xdx

Juan Spiller · Accepted Answer

Step 1
The given integral is as follows.

\int_{2}^{7} e^{2 - x} d x

Let u=2-x. Then, du=-dx.
The Lower limit will be u=2-2=0 and the upper limit will be u=2-7=-5.
Step 2
Apply the substitutions and evaluate the integral as follows.

\int_{2}^{7} e^{2 - x} d x = \int_{0}^{- 5} e^{u} (- d u)

= \int_{0}^{- 5} - e^{u} d u

= - {[e^{u}]}_{0}^{- 5}

= - [e^{- 5} - e^{0}]

= 1 - e^{- 5}

= 1 - \frac{1}{e^{5}}

Therefore, the value of the given integral is,

1 - \frac{1}{e^{5}}

yotaniwc · Accepted Answer

∫27e2−xdx  Evaluate the indefinite integral  ∫e2−xdx  ∫−etdt  −∫etdt  −et  −e2−x  Return the limits  −e2−x∣27  −e2−7−(−e2−2)  Simplify  Answer:  −1e5+1

Vasquez · Accepted Answer

∫e2−xdx This is a tabular integral: ∫e2−xdx=−e2−x+C Let&#039;s calculate a definite integral: ∫27e2−xdx=(−e2−x)|27 F(7)=−e−5 F(2)=−1 I=−e−5−(−1)=1−e−5

Evaluate the definite integral. \int_{2}^{7}e^{2-x}dx

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Answer & Explanation

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