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Question 1 of 5
Find the area bounded by the curve y = 4 − x 2 y = 4 − x 2 and lines x = 0 x = 0 and x = 1 x = 1
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Remember
Integrate the function using the power rule to find F ( x ) F ( x ) . Then to solve for the area between the curve and the axis, use the Definite Integral formula.
Identify the curve that the area is under and its bounds.
A A
= =
∫ ( 4 − x 2 ) d x ∫ ( 4 − x 2 ) d x
The curve is 4 - x 2 4 − x 2
A A
= =
∫ 1 0 ( 4 − x 2 ) d x ∫ 1 0 ( 4 − x 2 ) d x
The bounds are x = 0 x = 0 and x = 1 x = 1
Find the Indefinite Integral using the Power Rule
∫ ( 4 − x 2 ) d x ∫ ( 4 − x 2 ) d x
= =
4 ( x 0 + 1 0 + 1 ) – ( x 2 + 1 2 + 1 ) 4 ( x 0 + 1 0 + 1 ) – ( x 2 + 1 2 + 1 )
Apply the Power Rule
= =
4 ( x 1 1 ) – x 3 3 4 ( x 1 1 ) – x 3 3
Simplify
= =
4 x – x 3 3 4 x – x 3 3
Find the Definite Integral
A A
= =
∫ 1 0 ( 4 − x 2 ) d x ∫ 1 0 ( 4 − x 2 ) d x
= =
[ 4 x – x 3 3 ] 1 0 [ 4 x – x 3 3 ] 1 0
= =
[ 4 ( 1 ) - 1 3 3 ] – [ 4 ( 0 ) – 0 3 3 ] [ 4 ( 1 ) − 1 3 3 ] – [ 4 ( 0 ) – 0 3 3 ]
Substitute the upper ( 1 ) ( 1 ) and lower limits ( 0 ) ( 0 )
= =
[ 4 - 1 3 ] – [ 0 - 0 ] [ 4 − 1 3 ] – [ 0 − 0 ]
Simplify
= =
[ 12 3 - 1 3 ] – 0 [ 12 3 − 1 3 ] – 0
= =
11 3 11 3
11 3 square units 11 3 square units
Question 2 of 5
Find the area bounded by the curve y = 2 x 2 - 2 y = 2 x 2 − 2 and lines x = - 1 x = − 1 and x = 1 x = 1
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Remember
Integrate the function using the power rule to find F ( x ) F ( x ) . Then to solve for the area between the curve and the axis, use the Definite Integral formula.
Identify the curve that the area is under and its bounds.
A A
= =
∫ ( 2 x 2 − 2 ) d x ∫ ( 2 x 2 − 2 ) d x
The curve is 2 x 2 - 2 2 x 2 − 2
A A
= =
∫ 1 − 1 ( 2 x 2 − 2 ) d x ∫ 1 − 1 ( 2 x 2 − 2 ) d x
The bounds are x = - 1 x = − 1 and x = 1 x = 1
Find the Indefinite Integral using the Power Rule
∫ ( 2 x 2 − 2 ) d x ∫ ( 2 x 2 − 2 ) d x
= =
2 ( x 2 + 1 2 + 1 ) – 2 ( x 0 + 1 0 + 1 ) 2 ( x 2 + 1 2 + 1 ) – 2 ( x 0 + 1 0 + 1 )
Apply Power Rule
= =
2 ( x 3 3 ) – 2 ( x 1 1 ) 2 ( x 3 3 ) – 2 ( x 1 1 )
Simplify
= =
2 3 x 3 – 2 x 2 3 x 3 – 2 x
Find the Definite Integral
A A
= =
∫ 1 − 1 ( 2 x 2 − 2 ) d x ∫ 1 − 1 ( 2 x 2 − 2 ) d x
= =
[ 2 3 x 3 – 2 x ] 1 − 1 [ 2 3 x 3 – 2 x ] 1 − 1
= =
[ 2 3 ( 1 ) 3 - 2 ( 1 ) ] – [ 2 3 ( - 1 ) 3 – 2 ( - 1 ) ] [ 2 3 ( 1 ) 3 − 2 ( 1 ) ] – [ 2 3 ( − 1 ) 3 – 2 ( − 1 ) ]
Substitute the upper ( 1 ) ( 1 ) and lower limits ( - 1 ) ( − 1 )
= =
[ 2 3 ( 1 ) - 2 ] – [ 2 3 ( - 1 ) + 2 ] [ 2 3 ( 1 ) − 2 ] – [ 2 3 ( − 1 ) + 2 ]
Simplify
= =
[ 2 3 - 6 3 ] – [ - 2 3 + 6 3 ] [ 2 3 − 6 3 ] – [ − 2 3 + 6 3 ]
= =
- 4 3 – 4 3 − 4 3 – 4 3
= =
- 8 3 − 8 3
= =
8 3 8 3
Get the absolute value
8 3 square units 8 3 square units
Question 3 of 5
Find the area bounded by the curve y = x 2 - 7 x + 10 y = x 2 − 7 x + 10 and lines x = 2 x = 2 and x = 5 x = 5
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Remember
Integrate the function using the power rule to find F ( x ) F ( x ) . Then to solve for the area between the curve and the axis, use the Definite Integral formula.
Identify the curve that the area is under and its bounds.
A A
= =
∫ ( x 2 − 7 x + 10 ) d x ∫ ( x 2 − 7 x + 10 ) d x
The curve is x 2 - 7 x + 10 x 2 − 7 x + 10
A A
= =
∫ 5 2 ( x 2 − 7 x + 10 ) d x ∫ 5 2 ( x 2 − 7 x + 10 ) d x
The bounds are x = 2 x = 2 and x = 5 x = 5
Find the Indefinite Integral using the Power Rule
∫ ( x 2 − 7 x + 10 ) d x ∫ ( x 2 − 7 x + 10 ) d x
= =
( x 2 + 1 2 + 1 ) - 7 ( x 1 + 1 1 + 1 ) + 10 ( x 0 + 1 0 + 1 ) ( x 2 + 1 2 + 1 ) − 7 ( x 1 + 1 1 + 1 ) + 10 ( x 0 + 1 0 + 1 )
Apply Power Rule
= =
x 3 3 - 7 ( x 2 2 ) + 10 ( x 1 1 ) x 3 3 − 7 ( x 2 2 ) + 10 ( x 1 1 )
Simplify
= =
x 3 3 – 7 x 2 2 + 10 x x 3 3 – 7 x 2 2 + 10 x
Find the Definite Integral
A A
= =
∫ 5 2 ( x 2 − 7 x + 10 ) d x ∫ 5 2 ( x 2 − 7 x + 10 ) d x
= =
[ x 3 3 – 7 x 2 2 + 10 x ] 5 2 [ x 3 3 – 7 x 2 2 + 10 x ] 5 2
= =
[ 5 3 3 – 7 ( 5 2 2 ) + 10 ( 5 ) ] – [ 2 3 3 – 7 ( 2 2 2 ) + 10 ( 2 ) ] [ 5 3 3 – 7 ( 5 2 2 ) + 10 ( 5 ) ] – [ 2 3 3 – 7 ( 2 2 2 ) + 10 ( 2 ) ]
Substitute the upper ( 5 ) ( 5 ) and lower limits ( 2 ) ( 2 )
= =
[ 125 3 - 7 ( 25 2 ) + 50 ] – [ 8 3 - 7 ( 4 2 ) + 20 ] [ 125 3 − 7 ( 25 2 ) + 50 ] – [ 8 3 − 7 ( 4 2 ) + 20 ]
Simplify
= =
[ 125 3 - 175 2 + 50 ] – [ 8 3 - 14 + 20 ] [ 125 3 − 175 2 + 50 ] – [ 8 3 − 14 + 20 ]
= =
25 6 – 26 3 25 6 – 26 3
= =
- 9 2 − 9 2
= =
9 2 9 2
Get the absolute value
9 2 square units 9 2 square units
Question 4 of 5
Find the area bounded by the curve y = 1 2 x 3 + 1 y = 1 2 x 3 + 1 and lines x = 2 x = 2 and x = 0 x = 0
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Remember
Integrate the function using the power rule to find F ( x ) F ( x ) . Then to solve for the area between the curve and the axis, use the Definite Integral formula.
Identify the curve that the area is under and its bounds.
A A
= =
∫ ( 1 2 x 3 + 1 ) d x ∫ ( 1 2 x 3 + 1 ) d x
The curve is 1 2 x 3 + 1 1 2 x 3 + 1
A A
= =
∫ 2 0 ( 1 2 x 3 + 1 ) d x ∫ 2 0 ( 1 2 x 3 + 1 ) d x
The bounds are x = 0 x = 0 and x = 2 x = 2
Find the Indefinite Integral using the Power Rule
∫ ( 1 2 x 3 + 1 ) d x ∫ ( 1 2 x 3 + 1 ) d x
= =
1 2 ( x 3 + 1 3 + 1 ) + 1 ( x 0 + 1 0 + 1 ) 1 2 ( x 3 + 1 3 + 1 ) + 1 ( x 0 + 1 0 + 1 )
Apply Power Rule
= =
1 2 ( x 4 4 ) + x 1 2 ( x 4 4 ) + x
Simplify
= =
x 4 8 + x x 4 8 + x
Find the Definite Integral
A A
= =
∫ 2 0 ( 1 2 x 3 + 1 ) d x ∫ 2 0 ( 1 2 x 3 + 1 ) d x
= =
[ x 4 8 + x ] 2 0 [ x 4 8 + x ] 2 0
= =
[ 2 4 8 + 2 ] – [ 0 4 8 + 0 ] [ 2 4 8 + 2 ] – [ 0 4 8 + 0 ]
Substitute the upper ( 2 ) ( 2 ) and lower limits ( 0 ) ( 0 )
= =
[ 16 8 + 2 ] – [ 0 ] [ 16 8 + 2 ] – [ 0 ]
Simplify
= =
2 + 2 2 + 2
= =
4 4
4 square units 4 square units
Question 5 of 5
Find the area bounded by the curve y = x 3 y = x 3 and lines x = - 1 x = − 1 and x = 2 x = 2
Incorrect
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Remember
Integrate the function using the power rule to find F ( x ) F ( x ) . Then to solve for the area between the curve and the axis, use the Definite Integral formula.
Identify the curve that the area is under and its bounds.
A A
= =
A 1 + A 2 A 1 + A 2
Divide the area into two
A A
= =
∫ 0 − 1 ( x 3 ) d x + ∫ 2 0 ( x 3 ) d x ∫ 0 − 1 ( x 3 ) d x + ∫ 2 0 ( x 3 ) d x
The bounds are x = - 1 x = − 1 and x = 2 x = 2
Find the Indefinite Integral using the Power Rule
∫ ( x 3 ) d x ∫ ( x 3 ) d x
= =
x 3 + 1 3 + 1 x 3 + 1 3 + 1
Apply Power Rule
= =
x 4 4 x 4 4
Simplify
Find the Definite Integral
A A
= =
∫ 0 − 1 ( x 3 ) d x + ∫ 2 0 ( x 3 ) d x ∫ 0 − 1 ( x 3 ) d x + ∫ 2 0 ( x 3 ) d x
= =
[ x 4 4 ] 0 − 1 + [ x 4 4 ] 2 0 [ x 4 4 ] 0 − 1 + [ x 4 4 ] 2 0
= =
[ ( 0 4 4 ) – ( ( - 1 ) 4 4 ) ] + [ ( 2 4 4 ) – ( 0 4 4 ) ] [ ( 0 4 4 ) – ( ( − 1 ) 4 4 ) ] + [ ( 2 4 4 ) – ( 0 4 4 ) ]
Substitute the upper and lower limits
= =
1 4 + 16 4
Simplify
=
17 4