Miscellaneous Exercise answers: Application of Integrals

Open NCERT p. 298Matches NCERT’s answer

(i) y = x², x = 1, x = 2 and x-axis

1. The region lies under y = x², above the x-axis, between x = 1 and x = 2.
2. Area = ∫ from 1 to 2 of x² dx.
3. ∫x² dx = x³/3; putting in the limits gives 8/3 − 1/3 = 7/3.

Answer7/3 square units

(ii) y = x⁴, x = 1, x = 5 and x-axis

1. The region lies under y = x⁴, above the x-axis, between x = 1 and x = 5.
2. Area = ∫ from 1 to 5 of x⁴ dx.
3. ∫x⁴ dx = x⁵/5; putting in the limits gives 3125/5 − 1/5 = 3124/5.

Answer3124/5 square units

Watch the lesson Slicing a region into thin strips, and adding the strips with an integral

Open NCERT p. 298Matches NCERT’s answer

1. Evaluate: ∫ from −6 to 0 of |x + 3| dx.
2. y = |x+3| equals −(x+3) when x < −3, and (x+3) when x ≥ −3.

3. x	y = |x+3|
   −6	3
   −3	0
   0	3

4. Plot these points and join them with two straight lines meeting at (−3, 0); the graph is a V opening upwards.
5. To evaluate the integral, split it at x = −3, where the formula for y changes: from −6 to −3, and from −3 to 0.
6. From −6 to −3: ∫ −(x+3) dx = 9/2.
7. From −3 to 0: ∫ (x+3) dx = 9/2.
8. Total = 9/2 + 9/2 = 9.

Answer9

Watch this explained “The near miss”, 16:44 into Regions below the axis, and why the sign has to be discarded before adding

Open NCERT p. 298Matches NCERT’s answer

1. sin x is positive for x between 0 and π, and negative for x between π and 2π.
2. Area of first piece = ∫ from 0 to π of sin x dx = 2.
3. Area of second piece = size of ∫ from π to 2π of sin x dx = size of −2 = 2.
4. Total area = 2 + 2 = 4.

Answer4 square units

Watch this explained “The same lesson, moved along”, 13:32 into Regions below the axis, and why the sign has to be discarded before adding

Open NCERT p. 298Checked by computer

1. x³ is negative for x between −2 and 0, and positive for x between 0 and 1.
2. Area of first piece = size of ∫ from −2 to 0 of x³ dx = size of −4 = 4.
3. Area of second piece = ∫ from 0 to 1 of x³ dx = 1/4.
4. Total area = 4 + 1/4 = 17/4.

Answer17/4 square units, option (D).

Watch this explained “Four answers offered”, 14:18 into Regions below the axis, and why the sign has to be discarded before adding

Open NCERT p. 298Checked by computer

1. For x < 0, y = x|x| = −x²; for x > 0, y = x|x| = x².
2. Area of first piece = size of ∫ from −1 to 0 of −x² dx = size of −1/3 = 1/3.
3. Area of second piece = ∫ from 0 to 1 of x² dx = 1/3.
4. Total area = 1/3 + 1/3 = 2/3.

Answer2/3 square units, option (C).

Watch this explained “And one that goes further”, 15:33 into Regions below the axis, and why the sign has to be discarded before adding