Evaluate the integral. ∫01(x2+1)e−xdx

Question

Evaluate the integral.  ∫01(x2+1)e−xdx

xandir307dc · Accepted Answer

Step 1: Given that:
Evaluate the integral.

\int_{0}^{1} (x^{2} + 1) e^{- x} d x

Step 2: Formula Used:
Integration by Parts:

\int f (x) \cdot g (x) d x = f (x) \int g (x) d x - \int [\frac{d}{d x} (f (x)) \int g (x) d x] d x

Where f(x) is the first function and g(x) is second function.
Step 3: Evaluate:
To find the given definite integral first we find the value of the indefinite integral

\int (x^{2} + 1) e^{- x} d x = (x^{2} + 1) \int e^{- x} d x - \int [\frac{d}{d x} (x^{2} + 1) \int e^{- x} d x] d x

= - (x^{2} + 1) e^{- x} - \int 2 x (- e^{- x}) d x

= - (x^{2} + 1) e^{- x} + \int 2 x e^{- x} d x

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

= - (x^{2} + 1) e^{- x} + 2 (- x e^{- x} - e^{- x})

= - (x^{2} + 1) e^{- x} - 2 x e^{- x} - 2 e^{- x} + C

= - (x^{2} + 2 x + 3) e^{- x} + C

Now, Finding the definite integral for the given integral we have,

\int_{0}^{1} (x^{2} + 1) e^{- x} d x = {[- (x^{2} + 2 x + 3) e^{- x}]}_{0}^{1}

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Paineow · Accepted Answer

∫(x2+1)⋅e−xdx  Integration by parts formula:  ∫U(x)⋅dV(x)=U(x)⋅V(x)−∫V(x)⋅dU(x)  Put  U=x2+1  dV=e−xdx  Then:  dU=2x dx  V=−e−x  Therefore  ∫(x2+1)⋅e−x⋅dx=−(x2+1)⋅e−x−∫−2⋅x⋅e−x⋅dx=−(x2+1)⋅e−x+∫2⋅x⋅e−x⋅dx  Find the integral  ∫2⋅x⋅e−xdx  U=x  dU=dx  dV=2e−xdx  V=−2e−x  ∫2⋅x⋅e−xdx=−2⋅x⋅e−x−∫−2⋅e−xdx=−2⋅x⋅e−x+∫2⋅e−xdx  Find the integral  ∫2⋅e−xdx=−2⋅e−x  Answer:  ∫(x2+1)⋅e−x=−2⋅x⋅e−x+(−x2−1)⋅e−x−2⋅e−x+C  or  ∫(x2+1)⋅e−x=−(x2+2⋅x+3)⋅e−x+C  Calculate the definite integral:  ∫01(x2+1)⋅e−xdx=(−(x2+2⋅x+3)⋅e−x)∣01

karton · Accepted Answer

∫01(x2+1)e−xdx∫(x2+1)e−xdx∫(x2+1)×1exdx∫x2+1exdx∫x2ex+1exdx∫x2exdx+∫1exdx−x2e−x−2xe−x−2e−x−1ex(−x2e−x−2xe−x−2e−x−1ex)|01−12e−1−2×1e−1−2e−1−1e1−(−02e−0−2×0e−0−2e−0−1e0)−6e+3

Evaluate the integral. \int_{0}^{1}(x^{2}+1)e^{-x}dx

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