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katiebonar13
29.01.2021 •
Mathematics
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
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Ответ:
a.![\frac{3}{2}](/tpl/images/1076/8213/dbf42.png)
b.![\frac{15\pi }{4}](/tpl/images/1076/8213/a6ee4.png)
c.![\frac{81\pi }{2}](/tpl/images/1076/8213/0eed9.png)
d.![\frac{7}{2}](/tpl/images/1076/8213/447a9.png)
Step-by-step explanation:
P.S - The exact question is -
a.![\int\limits^1_0 {3x} \, dx](/tpl/images/1076/8213/0e540.png)
y = 3x
Shape of the given integral = A Triangle
Area of Triangle =
=
×1×3 = ![\frac{3}{2}](/tpl/images/1076/8213/dbf42.png)
The graph of integral is as follows :
b.![\int\limits^\15 _0 {\sqrt{15 - y^{2} } } \, dy](/tpl/images/1076/8213/40707.png)
x =![\sqrt{15 - y^{2} }](/tpl/images/1076/8213/4592e.png)
⇒x² = 15 - y²
⇒x² + y² = 15
⇒x² + y² = (√15)²
Shape of the integral = Quarter circle of radius √15
Area of Quarter circle =
=
×
×r² =
×
×(√15)² = ![\frac{15\pi }{4}](/tpl/images/1076/8213/a6ee4.png)
The graph of the following integral is as follows :
c.![\int\limits^9_-9 {\sqrt{81 - x^{2} } } \, dx](/tpl/images/1076/8213/0b2bb.png)
y =![\sqrt{81 - x^{2} }](/tpl/images/1076/8213/85ecf.png)
⇒y² = 81 - x²
⇒y² + x² = 81
⇒y² + x² = 9²
Shape of the integral = Semi circle of radius 9
Area of semi circle =
=
×
×r² =
×
×9² = ![\frac{81\pi }{2}](/tpl/images/1076/8213/0eed9.png)
The graph of the following integral is as follows :
d.![\int\limits^7_0 {5 - \frac{y}{7} } \, dy](/tpl/images/1076/8213/8250e.png)
x = 5 -![\frac{y}{7}](/tpl/images/1076/8213/32a45.png)
⇒
= 5 - x
⇒y = 35 - 7x
At y = 0 , x = 5
At y = 7 , x = 4
Shape of the integral = A triangle
Area of Triangle =
×(5-4)×7 =
×(1)×7 = ![\frac{7}{2}](/tpl/images/1076/8213/447a9.png)
The graph of the integral is as follows :
Ответ: