Question

The following integrals calculate areas of regions in the xy-plane. Say what shape each integral gives the area of. Be specific; for example, if the shape is a rectangle, give the base and height of the rectangle. Include a sketch of the region, showing the variable of integration, an arbitrary slice, and any other relevant quantities.

a. ∫ 3xdx
b. ∫ √15-y^2dy
c. ∫ √81- x^2 dx
d. ∫ (5- y/7)dy

Answer 1

a.
`(3)/(2)`

b.
`(15\pi )/(4)`

c.
`(81\pi )/(2)`

d.
`(7)/(2)`

Explanation:

P.S - The exact question is -

a.
`\int\limits^1_0 {3x} \, dx`

y = 3x

Shape of the given integral = A Triangle

Area of Triangle =
`\int\limits^1_0 {3x} \, dx` =
`(1)/(2)`×1×3 =
`(3)/(2)`

The graph of integral is as follows :

b.
`\int\limits^\15 _0 {\sqrt{15 - y^(2) } } \, dy`

x =
`\sqrt{15 - y^(2) }`

⇒x² = 15 - y²

⇒x² + y² = 15

⇒x² + y² = (√15)²

Shape of the integral = Quarter circle of radius √15

Area of Quarter circle =
`\int\limits^\15 _0 {\sqrt{15 - y^(2) } } \, dy` =
`(1)/(4)`×
`\pi`×r² =
`(1)/(4)`×
`\pi`×(√15)² =
`(15\pi )/(4)`

The graph of the following integral is as follows :

c.
`\int\limits^9_-9 {\sqrt{81 - x^(2) } } \, dx`

y =
`\sqrt{81 - x^(2) }`

⇒y² = 81 - x²

⇒y² + x² = 81

⇒y² + x² = 9²

Shape of the integral = Semi circle of radius 9

Area of semi circle =
`\int\limits^9_-9 {\sqrt{81 - x^(2) } } \, dx` =
`(1)/(2)`×
`\pi`×r² =
`(1)/(2)`×
`\pi`×9² =
`(81\pi )/(2)`

The graph of the following integral is as follows :

d.
`\int\limits^7_0 {5 - (y)/(7) } \, dy`

x = 5 -
`(y)/(7)`

⇒
`(y)/(7)` = 5 - x

⇒y = 35 - 7x

At y = 0 , x = 5

At y = 7 , x = 4

Shape of the integral = A triangle

Area of Triangle =
`(1)/(2)`×(5-4)×7 =
`(1)/(2)`×(1)×7 =
`(7)/(2)`

The graph of the integral is as follows :