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Worked problem · Calculus II · Unit 3A

Repeated Integration By Parts: Worked Example and Solution

A complete integration by parts example with method choice, derivation, verification, and a common wrong approach.

1 problem8 min estimated timeIntermediate to advanced progression

Problem

  1. Evaluate ∫x2ex dx\int x^2e^x\,dx.

Complete worked solutions

Every problem has a source-matched answer and independently reviewed derivation.

01

Problem 1: Evaluate ∫x2ex dx\int x^2e^x\,dx.

Answer: ex(x2−2x+2)+Ce^x(x^2-2x+2)+C

Why this method: Integration by parts matches the mathematical structure before any algebraic cleanup.

  1. Take u=x2u=x^2 and dv=exdxdv=e^x dx: I=x2ex−2∫xexdxI=x^2e^x-2\int xe^x dx.
  2. Using ∫xexdx=ex(x−1)\int xe^x dx=e^x(x-1), I=ex(x2−2x+2)+CI=e^x(x^2-2x+2)+C.
  3. Therefore the result is ex(x2−2x+2)+Ce^x(x^2-2x+2)+C.

What is included

See a concise answer and full derivation for evaluate integral x squared e to the power (x) d x.

Skills assessed

  • Integration by parts

Prerequisites

  • antiderivatives
  • product rule
Repeated Integration By Parts instructional sequence
A complete integration by parts example with method choice, derivation, verification, and a common wrong approach. The numbered labels and written sequence preserve meaning without relying on color.
Long description

Read the diagram from top to bottom. Each numbered box names one decision or mathematical operation. Arrows show the required order; the text labels remain the complete interpretation in print, dark mode, and nonvisual reading.

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Common errors

  • Choosing dv that is harder to integrate or losing the minus sign in the formula.

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Supporting steps