Question

Evaluate the integral

\(\int\limits_0^1 {\left( {{x^2} + 1} \right){e^{ - x}}.dx} \)

Short answer

The value of the Integralis

This is obtained by applying the formula,

\(\int {f\left( x \right).g\left( x \right).dx = f\left( x \right)\int {g\left( x \right).dx - \int {\left\{ {\frac{d}{{dx}}f\left( x \right).\int {g\left( x \right)} } \right\}dx} } } \)

Step-by-step solution

Step 1: Applying the formula

Given integration is\(\int\limits_0^1 {\left( {{x^2} + 1} \right){e^{ - x}}.dx} \)

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

• \( - \left( {{x^2} + 1} \right){e^{ - x}} - \int {2x.\left( { - {e^{ - x}}} \right)dx} \)
• \( - \left( {{x^2} + 1} \right).{e^{ - x}} + 2\int {x.{e^{ - x}}.dx} \)

Simplifying the next integration

• \( - \left( {{x^2} + 1} \right){e^{ - x}} + 2\left( {x\int {{e^{ - x}}.dx - \int {\left\{ {\frac{d}{{dx}}.x} \right\}\int {{e^{ - x}}.dx} } } } \right)\)
• \( - \left( {{x^2} + 1} \right){e^{ - x}} + 2\left( {x\int {{e^{ - x}}.dx - \int {1.\left( { - {e^{ - x}}} \right)dx} } } \right)\)
• \( - \left( {{x^2} + 1} \right){e^{ - x}} - 2x.{e^{ - x}} + 2\int {{e^{ - x}}.dx} \)
• \( - \left( {{x^2} + 1} \right){e^{ - x}} - 2x.{e^{ - x}} - 2{e^{ - x}}\)
• \( - {e^{ - x}}\left( {\left( {{x^2} + 1} \right) + 2x + 2} \right)\)
• \( - {e^{ - x}}\left( {{x^2} + 2x + 3} \right)\)

Applying the limits to the obtained solutions

• \( - \left( {{e^{ - x}}\left( {{x^2} + 2x + 3} \right)} \right)_0^1\)
• \( - \left( {{e^{ - 1}}\left( {1 + 2 + 3} \right) - {e^{\left( { - 0} \right)}}\left( {0 + 0 + 3} \right)} \right)\)
• \( - 2.207 + 3\)
• \(0.793\)

Therefore, the value of the given integral is \(\int\limits_0^1 {\left( {{x^2} + 1} \right){e^{ - x}}.dx = 0.793} \)