Answer:
a. [tex]\frac{3}{2}[/tex]
b. [tex]\frac{15\pi }{4}[/tex]
c. [tex]\frac{81\pi }{2}[/tex]
d. [tex]\frac{7}{2}[/tex]
Step-by-step explanation:
P.S - The exact question is -
a. [tex]\int\limits^1_0 {3x} \, dx[/tex]
y = 3x
Shape of the given integral = A Triangle
Area of Triangle = [tex]\int\limits^1_0 {3x} \, dx[/tex] = [tex]\frac{1}{2}[/tex]×1×3 = [tex]\frac{3}{2}[/tex]
The graph of integral is as follows :
b.[tex]\int\limits^\15 _0 {\sqrt{15 - y^{2} } } \, dy[/tex]
x = [tex]\sqrt{15 - y^{2} }[/tex]
⇒x² = 15 - y²
⇒x² + y² = 15
⇒x² + y² = (√15)²
Shape of the integral = Quarter circle of radius √15
Area of Quarter circle = [tex]\int\limits^\15 _0 {\sqrt{15 - y^{2} } } \, dy[/tex] = [tex]\frac{1}{4}[/tex]×[tex]\pi[/tex]×r² = [tex]\frac{1}{4}[/tex]×[tex]\pi[/tex]×(√15)² = [tex]\frac{15\pi }{4}[/tex]
The graph of the following integral is as follows :
c.[tex]\int\limits^9_-9 {\sqrt{81 - x^{2} } } \, dx[/tex]
y = [tex]\sqrt{81 - x^{2} }[/tex]
⇒y² = 81 - x²
⇒y² + x² = 81
⇒y² + x² = 9²
Shape of the integral = Semi circle of radius 9
Area of semi circle = [tex]\int\limits^9_-9 {\sqrt{81 - x^{2} } } \, dx[/tex] = [tex]\frac{1}{2}[/tex]×[tex]\pi[/tex]×r² = [tex]\frac{1}{2}[/tex]×[tex]\pi[/tex]×9² = [tex]\frac{81\pi }{2}[/tex]
The graph of the following integral is as follows :
d. [tex]\int\limits^7_0 {5 - \frac{y}{7} } \, dy[/tex]
x = 5 - [tex]\frac{y}{7}[/tex]
⇒[tex]\frac{y}{7}[/tex] = 5 - x
⇒y = 35 - 7x
At y = 0 , x = 5
At y = 7 , x = 4
Shape of the integral = A triangle
Area of Triangle = [tex]\frac{1}{2}[/tex]×(5-4)×7 = [tex]\frac{1}{2}[/tex]×(1)×7 = [tex]\frac{7}{2}[/tex]
The graph of the integral is as follows :
