Area under a curve: AS mathematics

1. FINDING THE AREA
UNDER A CURVE
USING INTEGRATION
AS P1 MATH
BY: WILLIAM, VINCENT, WENDY

2. INTRODUCTION
• INTEGRATION IS ALSO 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
• 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

6. USE
INTEGRATION
TO FIND AREA
THE AREA 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 THE AREA UNDER THE CURVE 𝑦 = 𝑥2 + 2 IN WHICH THE AREA IS BOUNDED
BETWEEN X=2 AND X=6.

8. SOLUTION
𝐴 = 𝑎
𝑏
𝑓 𝑥 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 THE X-AXIS
• IF A CURVE LIES BELOW THE X-AXIS, THE AREA BETWEEN THAT PART AND THE
X-AXIS IS
− 𝑦 𝑑𝑥

10. EXAMPLE
• THE CURVE 𝑦 = 𝑥2 − 3𝑥 + 2 = (𝑥 − 2)(𝑥 − 1) MEET THE X-AXIS WHERE X=1 AND
X=2.
FIND THE SHADED AREA.

11. 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.

12. ANOTHER TYPE OF CURVE…
• 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 A CURVE 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, MAKE X 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. QUESTION???