Question 26 of 33intermediate🔧 ApplyShort Answer2 marks

Integrate the functions in Exercises 1 to 22: x2x^{2} exe^{x}

Correct Answer

exe^{x}(x2x^{2} - 2x + 2) + C

Exercise: EXERCISE 7.6 | Q: 3 | (Page 39)
For More Understanding

Explanation

This question from Exercise 7.6 requires integration by parts. Since x² is an algebraic function and exe^{x} is an exponential function, we take x² as the first function and exe^{x} as the second function (following ILATE rule). Integration by parts needs to be applied twice because the power of x is 2.

Solution Steps

  1. Step 1: Apply integration by parts with u = x² and v = exe^{x}. ∫x²exe^{x} dx = x²·exe^{x} - ∫2x·exe^{x} dx = x²exe^{x} - 2∫xexxe^{x} dx

  2. Step 2: Apply integration by parts again on ∫xexxe^{x} dx with u = x and v = exe^{x}. ∫xexxe^{x} dx = x·exe^{x} - ∫exe^{x} dx = xexxe^{x} - exe^{x} + C

  3. Step 3: Substitute back: ∫x²exe^{x} dx = x²exe^{x} - 2(xexxe^{x} - exe^{x}) + C = x²exe^{x} - 2xex2xe^{x} + 2ex2e^{x} + C = exe^{x}(x² - 2x + 2) + C

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