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![DEFINITE INTEGRATION
• WHEN WE ARE GIVEN VALUES LIKE THIS: 𝑎
𝑏
𝑓 𝑥 𝑑𝑥 , WHERE A AND B ARE THE
LIMITS OF THE INTEGRAL, THIS IS KNOWN AS DEFINITE INTEGRALS.
• EXAMPLE : −2
0
1 − 𝑡 − 𝑡2 𝑑𝑡
= 0 − [−2 −
−2 2
2
−
−2 3
3
]
=0 − −1
1
3
= 1
1
3](https://image.slidesharecdn.com/findingtheareaunderacurveusingintegration-150715113309-lva1-app6892/85/Finding-the-area-under-a-curve-using-integration-5-320.jpg)
![SOLUTION
• 𝑦 𝑑𝑥 WILL HAVE A NEGATIVE DOMAIN.
• REMEMBER 𝑎
𝑏
𝑓 𝑥 𝑑𝑥
• 1
2
𝑦 𝑑𝑥 = 1
2
𝑥2 − 3𝑥 + 2 𝑑𝑥 = [
𝑥3
3
−
3𝑥2
2
+ 2𝑥]
= (
8
3
− 6 + 4) − (
1
3
−
3
2
+ 2)
=
2
3
−
5
6
= −
1
6
THIS IS THE AREA BELOW THE X-AXIS.](https://image.slidesharecdn.com/findingtheareaunderacurveusingintegration-150715113309-lva1-app6892/85/Finding-the-area-under-a-curve-using-integration-11-320.jpg)
Uploaded byChristopher Chibangu
Finding the area under a curve using integration
This document discusses using integration to find the area under a curve. It defines integration as the reverse of differentiation and shows how to find antiderivatives. It provides examples of indefinite and definite integrals, and explains how definite integrals between limits can be used to calculate the area under a curve over an interval. The document demonstrates this by finding the area under various curves bounded between given x-values. It also discusses how to handle areas below or partly above and below the x-axis.
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Finding the area under a curve using integration
- 1. FINDING THE AREA UNDERA CURVE USING INTEGRATION AS P1 MATH BY: WILLIAM, VINCENT, WENDY
- 2. INTRODUCTION • INTEGRATION ISALSO CALLED ANTI-DIFFERENTIATION. THIS MEANS THAT IT IS REVERSE DIFFERENTIATION. • IF 𝒅𝒚 𝒅𝒙 = 𝒂𝒙 𝒏 THEN 𝒚 = 𝒂𝒙 𝒏+𝟏 𝒏+𝟏 , WHERE 𝒏 ≠ −𝟏.
- 3. EXAMPLE 1 • INTEGRATE: •SOLUTION: 𝑑𝑦 𝑑𝑥 = 2𝑥3 − 3𝑥 + 1 𝑑𝑦 𝑑𝑥 = (2𝑥3 − 3𝑥 + 1 )𝑑𝑥 𝑦 = 2𝑥3+1 3 + 1 − 3𝑥1+1 1 + 1 + 1𝑥0+1 0 + 1 = 𝟏 𝟐 𝒙 𝟒 − 𝟑 𝟐 𝒙 𝟐 + 𝒙 + 𝒌 2𝑥3 − 3x + 1
- 4. EXAMPLE 2 • INTEGRATE: •SOLUTION: EXPAND 2𝑥 − 3 2 = 4X2 − 12X + 9 𝑑𝑦 𝑑𝑥 = 4𝑥2 − 12𝑥 + 9 𝑑𝑥 = 4𝑥2+1 2 + 1 − 12𝑥1+1 1 + 1 + 9𝑥0+1 0 + 1 + 𝑘 𝒚 = 𝟒𝒙 𝟑 𝟑 − 𝟔𝒙 𝟐 + 𝟗𝒙 + 𝒌 2𝑥 − 3 2
- 5. DEFINITE INTEGRATION • WHENWE ARE GIVEN VALUES LIKE THIS: 𝑎 𝑏 𝑓 𝑥 𝑑𝑥 , WHERE A AND B ARE THE LIMITS OF THE INTEGRAL, THIS IS KNOWN AS DEFINITE INTEGRALS. • EXAMPLE : −2 0 1 − 𝑡 − 𝑡2 𝑑𝑡 = 0 − [−2 − −2 2 2 − −2 3 3 ] =0 − −1 1 3 = 1 1 3
- 6. USE INTEGRATION TO FIND AREA THEAREA UNDER A GRAPH CAN BE FOUND BY USING THE FORMULA… WHERE A IS THE LOWER LIMIT AND B IS THE UPPER LIMIT.
- 7. EXAMPLE • FIND THEAREA UNDER THE CURVE 𝑦 = 𝑥2 + 2 IN WHICH THE AREA IS BOUNDED BETWEEN X=2 AND X=6.
- 8. SOLUTION 𝐴 = 𝑎 𝑏 𝑓𝑥 𝑑𝑥 X = 2 X = 6 𝐴 = 2 6 𝑥2 + 2 𝑑𝑥 = 2 6 𝑥2+1 2 + 1 + 2𝑥0+1 0 + 1 𝐴 = 2 6 𝑥3 3 + 2𝑥 1 = 2 6 1 3 𝑥3 + 2𝑥 𝐴 = 1 3 6 3 + 2 6 − 1 3 2 3 + 2 2 𝐴 = 72 + 12 − 8 3 + 4 = 77.3
- 9. AREA UNDER THEX-AXIS • IF A CURVE LIES BELOW THE X-AXIS, THE AREA BETWEEN THAT PART AND THE X-AXIS IS − 𝑦 𝑑𝑥
- 10. EXAMPLE • THE CURVE𝑦 = 𝑥2 − 3𝑥 + 2 = (𝑥 − 2)(𝑥 − 1) MEET THE X-AXIS WHERE X=1 AND X=2. FIND THE SHADED AREA.
- 11. SOLUTION • 𝑦 𝑑𝑥WILL HAVE A NEGATIVE DOMAIN. • REMEMBER 𝑎 𝑏 𝑓 𝑥 𝑑𝑥 • 1 2 𝑦 𝑑𝑥 = 1 2 𝑥2 − 3𝑥 + 2 𝑑𝑥 = [ 𝑥3 3 − 3𝑥2 2 + 2𝑥] = ( 8 3 − 6 + 4) − ( 1 3 − 3 2 + 2) = 2 3 − 5 6 = − 1 6 THIS IS THE AREA BELOW THE X-AXIS.
- 12. ANOTHER TYPE OFCURVE… • IF A CURVE LIES PARTLY ABOVE AND PARTLY BELOW THE X-AXIS, THE TOTAL AREA WILL BE 𝑎 𝑏 𝑦 𝑑𝑥 − 𝑏 𝑐 𝑦 𝑑𝑥
- 13. AREA BETWEEN ACURVE AND THE Y-AXIS • THE GRAPH SHOWS THE PART OF THE CURVE 𝑦 − 1 2 = 𝑥 − 1 , FIND THE AREA OF THE SHADED REGION.
- 14. SOLUTION • FIRSTLY, MAKEX THE SUBJECT OF THE EQUATION OF THE CURVE. 𝑥 = 𝑦 − 1 2 + 1 = 𝑦2 + 2𝑦 + 2 FIND THE AREA 0 1 𝑦2 + 2𝑦 + 2 𝑑𝑦 = 𝑦3 3 − 𝑦2 + 2𝑦 0 1 = 1 3 − 1 + 2 − 0 = 1 1 3 𝑢𝑛𝑖𝑡𝑠2
- 15.