Formulae [2]
Formula: Modulus Functions
Break modulus into cases:
\[|x-a|=
\begin{cases}
x-a, & x\geq a \\
a-x, & x<a & &
\end{cases}\]
Formula: Area between Two Curves
\[\text{Area between two curves}=\int_a^b[\text{upper curve - lower curve}]dx\]
\[\text{Area between two curves}=\int_{a}^{b}y\mathrm{of}f(x)dx-\int_{a}^{b}y\mathrm{of}g(x)dx\]
Key Points
Key Points: Geometrical Interpretation of Definite Integral
The area bounded by the curve y = f (x), the x-axis and the ordinates. x = a, x=b is \[\int_a^bydx\].
Sign of Area:
| Condition | Result |
|---|---|
| Curve above the x-axis | Area is positive |
| Curve below the x-axis | Area is negative |
| Curve cuts the x-axis | Integral ≠ actual area |
Key Points: Symmetry
| Type of Symmetry | What to Replace | Condition | Result |
|---|---|---|---|
| About y-axis | Replace (x) by (-x) | Equation unchanged | Symmetrical about the y-axis |
| About x-axis | Replace (y) by (-y) | Equation unchanged | Symmetrical about the x-axis |
| About origin | Replace (x) by (-x), (y) by (-y) | Equation unchanged | Symmetrical about the origin |
| About y = x | Interchange (x) and (y) | Equation unchanged | Symmetrical about the line y = x |
| About y = −x | Replace (x) by (-y), (y) by (-x) | Equation unchanged | Symmetrical about the line y = −x |
Key Points: Area Under a Curve
-
If the curve is ,
\[\mathrm{Area}=\int_a^bf(x)dx\] -
If curve is x = g(y),
\[\mathrm{Area}=\int_c^dg(y)dy\] -
If the curve is on both sides → split + add
When to Use:
| Curve form | Formula |
|---|---|
| y = f(x) | \[\int ydx\] |
| x = f(y) | \[\int xdy\] |
Key Points: Standard Curves
| Curve | Shape |
|---|---|
| \[y=\sqrt{a^2-x^2}\] | Upper semicircle |
| \[x^2+y^2=a^2\] | Circle |
| \[y^2=4ax\] | Right parabola |
| \[x^2=4ay\] | Upward parabola |
| y = sin x,cos x | Wave (sign changes!) |
Important Questions [37]
- Using integration, find the area of the region bounded by y = mx (m > 0), x = 1, x = 2 and the X-axis.
- Using integration, find the area bounded by the curve x^2 = 4y and the line x = 4y − 2
- Prove that the curves y^2 = 4x and x^2 = 4y divide the area of square bounded by x = 0, x = 4, y = 4 and y = 0 into three equal parts.
- Using integration, find the area of the region bounded by the lines y = 2 + x, y = 2 – x and x = 2.
- Using the Method of Integration, Find the Area of the Triangular Region Whose Vertices Are (2, -2), (4, 3) and (1, 2).
- Sketch the region bounded by the curves y=√(5-x^2) and y=|x-1| and find its area using integration.
- Using Integration, Find the Area of the Region Bounded by the Line X – Y + 2 = 0, the Curve X = √ Y and Y-axis.
- Using integration, find the area of the region bounded by the parabola y2 = 4x and the circle 4x2 + 4y2 = 9.
- Using the method of integration, find the area of the region bounded by the lines 3x − 2y + 1 = 0, 2x + 3y − 21 = 0 and x − 5y + 9 = 0
- Using integration, find the area of the smaller region bounded by the ellipse "x"^2/9+"y"^2/4=1and the line "x"/3+"y"/2=1.
- Find the area bounded by the curve y = |x – 1| and y = 1, using integration.
- Find the area of the region bounded by curve 4x2 = y and the line y = 8x + 12, using integration.
- Using integration, find the area of the region bounded by the curves x2 + y2 = 4, x = 3y and x-axis lying in the first quadrant.
- Find the area of the region enclosed by the curves y2 = x, x = 14, y = 0 and x = 1, using integration.
- Using integration, find the area of the region bounded by line y = 3x, the curve y = 4-x2 and Y-axis in first quadrant.
- Find the area of the smaller region bounded by the curves x225+y216 = 1 and x5+y4 = 1, using integration.
- Sketch the region bounded by the lines 2x + y = 8, y = 2, y = 4 and the Y-axis. Hence, obtain its area using integration.
- Find the area of the following region using integration ((x, y) : y2 ≤ 2x and y ≥ x – 4).
- Find the area of the minor segment of the circle x2 + y2 = 4 cut off by the line x = 1, using integration.
- Find the area, lying above x-axis and included between the circle x2 + y2 = 8x and the parabola y2 = 4x.
- Find the Area of the Region in the First Quadrant Enclosed by the X-axis, the Line Y = X and the Circle X2 + Y2 = 32.
- Using Integration Find the Area of the Region Bounded by the Curves Y = √ 4 − X 2 , X 2 + Y 2 − 4 X = 0 and the X-axis.
- Using the Method of Integration, Find the Area of the Triangle Abc, Coordinates of Whose Vertices Area A(1, 2), B (2, 0) and C (4, 3).
- Find the Coordinates of a Point of the Parabola Y = X2 + 7x + 2 Which is Closest to the Straight Line Y = 3x − 3.
- Using integration, find the area of the region bounded by the triangle whose vertices are (−1, 2), (1, 5) and (3, 4).
- Using Integration, Find the Area of Region Bounded by the Triangle Whose Vertices Are (–2, 1), (0, 4) and (2, 3).
- Using Integration, find the area of triangle whose vertices are (– 1, 1), (0, 5) and (3, 2).
- Find the Area Enclosed Between the Parabola 4y = 3x2 and the Straight Line 3x - 2y + 12 = 0.
- Using integration find the area of the region {(x, y) : x2+y2⩽ 2ax, y2⩾ ax, x, y ⩾ 0}.
- Find the area of the region in the first quadrant enclosed by the x-axis, the line y = x and the circle x^2 + y^2 = 32.
- Find the Area of the Smaller Region Bounded by the Ellipse X 2 9 + Y 2 4 = 1 and the Line X 3 + Y 2 = 1 .
- Find the Area of the Region. {(X,Y) : 0 ≤ Y ≤ X2 , 0 ≤ Y ≤ X + 2 ,-1 ≤ X ≤ 3} .
- Using Integration Find the Area of the Triangle Formed by Negative X-axis and Tangent and Normal to the Circle X 2 + Y 2 = 9 at ( − 1 , 2 √ 2 ) .
- Using Integration, Find the Area of the Region {(X, Y) : X2 + Y2 ≤ 1 ≤ X + Y}.
- Find the Area Bounded by the Circle X2 + Y2 = 16 and the Line `Squareroot 3 Y = X` in the First Quadrant, Using Integration.
- Using integration find the area of the triangle formed by positive x-axis and tangent and normal of the circle
- Using the Method of Integration, Find the Area of the Triangle Abc, Coordinates of Whose Vertices Are a (4 , 1), B (6, 6) and C (8, 4).
