Chapter 17: Problem 10

Find \(\int_{0}^{1}\left(x^{2}-3 x+1\right) \mathrm{e}^{x} \mathrm{~d} x\).

Short Answer

Expert verified

Question: Find the definite integral \(\int_{0}^{1}\left(x^{2}-3 x+1\right) e^{x} dx\). Answer: The definite integral is \(-e+1\).

Step by step solution

Use integration by parts formula

To use integration by parts, we first need to choose which parts of the given function to integrate and differentiate. Since we want a simpler function for our integral, let's differentiate \(x^2-3x+1\) and integrate \(e^x\). The integration by parts formula is:\[ \int u\, dv = uv - \int v\, du\]Let \(u = x^2-3x+1\) and \(dv = e^x dx\).

Differentiate u and integrate dv

Next, differentiate \(u\) with respect to \(x\), and integrate \(dv\): \[du = (2x - 3) dx\] \[v = \int e^x dx = e^x\]

Apply integration by parts formula

Now, we plug our \(u\), \(v\), and \(du\) into the integration by parts formula: \[\int_{0}^{1}(x^2 - 3x + 1)e^x dx = \int_{0}^1 u\, dv = uv \Big|_0^1 - \int_{0}^{1} v\, du\] \[= (x^2 - 3x + 1)e^x\Big|_0^1 - \int_{0}^{1} (2x - 3)e^x dx\]

Compute uv from limits

First, compute the value of \(uv\) at the limits 0 and 1: \[((1^2 - 3(1) + 1)e^1 - (0^2 - 3(0) + 1)e^0) = (1 - 3 + 1)e - (1) = (1 - 2)(e - 1)\]

Integrate using integration by parts formula again

Now, apply the integration by parts formula again for the integral in step 3. Let \(u = 2x - 3\) and \(dv = e^x dx\): \[du = 2 dx\] \[v = \int e^x dx = e^x\] The integral becomes: \[-\int_{0}^{1} (2x - 3)e^x dx = -\int_{0}^1 u\, dv = -\left[uv \Big|_0^1 - \int_{0}^{1} v\, du\right]\]

Compute new uv from limits

Compute the value of \(uv\) at the limits 0 and 1: \[ (2(1)-3)e^1 - (2(0)-3)e^0 = -e - 3\]

Integrate v du

Now we need to integrate \(v \, du = 2e^x dx\). This integral has a straightforward antiderivative: \[\int_{0}^{1} 2e^x dx = 2\int_{0}^{1} e^x dx = 2[e^x\Big|_0^1] = 2(e - 1)\]

Complete the integration by parts computation

In step 5, we found an expression for the integration by parts formula for the integral in step 3. We found the values of two separate components: \[uv \Big|_0^1 = -e - 3\] \[\int_{0}^{1} v\, du = 2(e - 1)\] So, the integral from step 3 becomes: \[-\left[-e - 3 - 2(e - 1)\right] = e + 3 - 2e + 2 = 5 - e\]

Combine results to find definite integral

Now that we have the results from steps 4 and 8, we can combine them to get our final answer: \[\int_{0}^{1}(x^2 - 3x + 1)e^x dx = (1 - 2)(e - 1) - (5 - e)\] \[ = -2e + 2 + e - 1 = -e + 1\] Therefore, the definite integral is \(\boxed{-e + 1}\).

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Find \(\int_{0}^{1}\left(x^{2}-3 x+1\right) \mathrm{e}^{x} \mathrm{~d} x\).

Short Answer

Step by step solution

Use integration by parts formula

Differentiate u and integrate dv

Apply integration by parts formula

Compute uv from limits

Integrate using integration by parts formula again

Compute new uv from limits

Integrate v du

Complete the integration by parts computation

Combine results to find definite integral

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