---
title: "Double integral xy(x+y) between curves"
subject: "Mathematics"
chapter: 2
key: "semester_2/mathematics:double-integral-xy-x-y"
source: https://www.bctengineer.com/q/semester_2/mathematics/double-integral-xy-x-y
site: BCT Engineering
---

# Double integral xy(x+y) between curves

Mathematics · Chapter 2: Multiple Integrals

- Asked in: 2074 Bhadra, 5 marks; 2073 Bhadra, 5 marks; 2073 Magh, 5 marks; 2069 Bhadra, 5 marks (asked 4 times)
- Video: [VTU Engineering Maths 1 Evaluate ∬xy(x+y)dydx taken over the area between y=x^2 & y=x (PART-3)](https://www.youtube.com/watch?v=kdnBmsia8Qk) — EASY MATHS EASY TRICKS
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/mathematics/double-integral-xy-x-y
- Whole chapter: https://www.bctengineer.com/md/semester_2/mathematics/chapter-2

## Question

Evaluate ∬ xy(x+y) dx dy over the area between y = x² and y = x.

## Short answer

Integrate over region between y=x² and y=x, 0≤x≤1. Inner integral w.r.t. y: ∫_{x²}^{x} (x²y+xy²) dy = (5/6)x⁴ − x⁶/2 − x⁷/3. Outer integration yields I = 3/56.

## Answer

**Region.** The curves y = x² and y = x meet where x² = x, i.e. at x = 0 and x = 1 (points (0,0) and (1,1)). Between them, x² ≤ y ≤ x for 0 ≤ x ≤ 1.

I = ∫₀¹ ∫_{x²}^{x} xy(x + y) dy dx = ∫₀¹ ∫_{x²}^{x} (x²y + xy²) dy dx

#### Inner integral (x constant)
```text
∫ (x²y + xy²) dy = x² y²/2 + x y³/3

at y = x :   x⁴/2 + x⁴/3
at y = x² :  x⁶/2 + x⁷/3

inner = (x⁴/2 − x⁶/2) + (x⁴/3 − x⁷/3)
      = (5/6)x⁴ − x⁶/2 − x⁷/3
```

#### Outer integral
```text
I = ∫₀¹ [ (5/6)x⁴ − x⁶/2 − x⁷/3 ] dx
  = (5/6)(1/5) − (1/2)(1/7) − (1/3)(1/8)
  = 1/6 − 1/14 − 1/24
  = 9/168 = 3/56
```

Using common denominator 168: 1/6 = 28/168, 1/14 = 12/168, 1/24 = 7/168, so I = (28 − 12 − 7)/168 = 9/168 = 3/56.

**Answer: I = 3/56.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Curves y = x² and y = x meet at:
   A. x = 1 only
   B. x = −1 and x = 1
   C. x = 0 and x = 2
   D. x = 0 and x = 1
   Answer: D. x² = x. (Key point: Limits)

2. For 0 ≤ x ≤ 1, y ranges:
   A. from x to x²
   B. from x² to x
   C. from 0 to 1
   D. from 0 to x²
   Answer: B. x² ≤ x on [0,1], so the parabola is lower. (Key point: Order)

3. The value of the integral is:
   A. 9/56
   B. 1/6
   C. 3/56
   D. 1/14
   Answer: C. 1/6 − 1/14 − 1/24 = 9/168 = 3/56. (Key point: Result)

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