Miscellaneous Exercise — Application of Integrals

1. 13 marksNCERT Exercise

   Find the area under the given curves and given lines: (i) y=x^2, x=1, x=2 and x-axis (ii) y=x^4, x=1, x=5 and x-axis.

   Hint. Both curves are positive throughout their given intervals, so no sign-change split is needed — just integrate directly and apply the power rule.

   For (i), since x^2 is positive on [1,2], the area is integral from 1 to 2 of x^2 dx = [x^3/3] from 1 to 2 = 8/3-1/3 = 7/3. For (ii), since x^4 is positive on [1,5], the area is integral from 1 to 5 of x^4 dx = [x^5/5] from 1 to 5 = 3125/5-1/5 = 3124/5.

   ✦ (i) 7/3 square units. (ii) 3124/5 square units

2. 23 marksNCERT Exercise

   Sketch the graph of y=|x+3| and evaluate integral from -6 to 0 of |x+3| dx.

   Hint. The expression inside the absolute value changes sign at x=-3, which lies inside the interval [-6,0], so split the integral there.

   Since x+3 is negative on [-6,-3] and positive on [-3,0], split the integral: integral from -6 to -3 of -(x+3)dx + integral from -3 to 0 of (x+3)dx. Each piece evaluates to 9/2 by the power rule, since the graph is two mirror-image triangles of equal area.

   ✦ 9 square units

3. 33 marksNCERT Exercise

   Find the area bounded by the curve y=sin x between x=0 and x=2*pi.

   Hint. sinx is positive on [0,pi] and negative on [pi,2pi], so split the integral at pi and take the absolute value of the second piece.

   Since sinx>=0 on [0,pi] and sinx<=0 on [pi,2pi], the area is integral from 0 to pi of sinx dx + |integral from pi to 2pi of sinx dx| = 2+2, because each half-period of sine contributes exactly 2 to the area by symmetry.

   ✦ 4 square units

4. 43 marksNCERT Exercise

   Choose the correct answer: area bounded by the curve y=x^3, the x-axis, and the ordinates x=-2 and x=1 is (A) -9 (B) -15/4 (C) 15/4 (D) 17/4

   Hint. x^3 changes sign at x=0, which lies inside [-2,1], so split the integral there and take the absolute value of the negative piece.

   Since x^3<0 on [-2,0] and x^3>0 on [0,1], the area is |integral from -2 to 0 of x^3 dx| + integral from 0 to 1 of x^3 dx = |-4| + 1/4, which gives the final sum.

   ✦ (D) 17/4

5. 53 marksNCERT Exercise

   Choose the correct answer: the area bounded by the curve y=x|x|, x-axis, and the ordinates x=-1 and x=1 is given by (A) 0 (B) 1/3 (C) 2/3 (D) 4/3 [Hint: y=x^2 if x>0 and y=-x^2 if x<0]

   Hint. Following the hint, split the integral at x=0 since the curve switches from -x^2 to x^2 there.

   Since y=-x^2 on [-1,0] and y=x^2 on [0,1], the area is |integral from -1 to 0 of -x^2 dx| + integral from 0 to 1 of x^2 dx = 1/3+1/3, so the two equal pieces combine to give the answer.

   ✦ (C) 2/3

Solutions written by the tuition.in editorial team and checked against lemh202.pdf (NCERT, Reprint 2026-27). Questions are referenced from the NCERT textbook for identification.