JEE Mathematics Area Under The Curve MCQs Set 08

Answer: (c) \( \frac{4}{3} \)

Explanation:
1. Put \( y = 0 \): \( x^2 - 4x + 3 = 0 \), so the curve meets the x-axis at \( x = 1 \) and \( x = 3 \).
2. Between these points the parabola lies above the x-axis (it opens downwards).
3. Area \( = \int_1^3 (4x - x^2 - 3)\,dx = \left[2x^2 - \frac{x^3}{3} - 3x\right]_1^3 = 0 - \left(-\frac{4}{3}\right) = \frac{4}{3} \).

Teacher's Note:
1. Always find the x-intercepts first; they give the limits of integration.
2. A parabola cutting the x-axis at \( \alpha, \beta \) with leading coefficient \( a \) encloses area \( \frac{|a|(\beta-\alpha)^3}{6} \), here \( \frac{8}{6} = \frac{4}{3} \).

Answer: (d) \( \frac{3}{2} \)

Explanation:
1. The line \( y = x + 1 \) meets the x-axis at \( (-1, 0) \) and the y-axis at \( (0, 1) \); the curve \( y = \cos x \) also passes through \( (0, 1) \) and reaches the x-axis at \( x = \frac{\pi}{2} \).
2. For \( -1 \le x \le 0 \) the region lies under the line: a right triangle of area \( \frac{1}{2} \).
3. For \( 0 \le x \le \frac{\pi}{2} \) the region lies under the cosine curve: \( \int_0^{\pi/2} \cos x\,dx = 1 \).
4. Total area \( = \frac{1}{2} + 1 = \frac{3}{2} \).

Teacher's Note:
1. When the upper boundary changes from one curve to another, split the area at the point where they meet.
2. A triangle under a straight line is quicker to find with geometry than with integration.

Answer: (c) \( \frac{81}{2} \)

Explanation:
1. Write the curve as \( x = 9y - y^3 \). It meets the y-axis (\( x = 0 \)) at \( y = 0, 3, -3 \).
2. The curve is symmetric about the origin, so the two loops (\( 0 \le y \le 3 \) and \( -3 \le y \le 0 \)) have equal area.
3. Area \( = 2\int_0^3 (9y - y^3)\,dy = 2\left[\frac{9y^2}{2} - \frac{y^4}{4}\right]_0^3 = 2\left(\frac{81}{2} - \frac{81}{4}\right) = \frac{81}{2} \).

Teacher's Note:
1. When x is given as a function of y, integrate with respect to y between the y-intercepts.
2. Symmetry about the origin lets you double one loop instead of handling a sign change.

Answer: (c) \( e^2 - 2 \)

Explanation:
1. The curve \( y = e^x \) meets \( y = 1 - x \) at \( x = 0 \) (point \( (0, 1) \)) and lies above the line \( y = |x - 1| \) for \( 0 \le x \le 2 \).
2. For \( 0 \le x \le 1 \), \( |x - 1| = 1 - x \), giving \( \int_0^1 (e^x - 1 + x)\,dx = e - \frac{3}{2} \).
3. For \( 1 \le x \le 2 \), \( |x - 1| = x - 1 \), giving \( \int_1^2 (e^x - x + 1)\,dx = e^2 - e - \frac{1}{2} \).
4. Adding: \( e - \frac{3}{2} + e^2 - e - \frac{1}{2} = e^2 - 2 \).

Teacher's Note:
1. Remove the modulus by splitting at the point where the expression inside becomes zero.
2. Check that the upper curve stays the same over each piece before subtracting.

Answer: (b) \( \frac{4 - 3\ln 3}{2} \)

Explanation:
1. For \( x < 2 \): \( y = x \); it meets \( y = \frac{3}{x} \) at \( x = \sqrt{3} \). For \( x > 2 \): \( y = 4 - x \); it meets \( y = \frac{3}{x} \) where \( x^2 - 4x + 3 = 0 \), i.e. \( x = 3 \) (the root \( x = 1 \) is not in this range).
2. Area \( = \int_{\sqrt{3}}^{2}\left(x - \frac{3}{x}\right)dx + \int_2^3\left(4 - x - \frac{3}{x}\right)dx \).
3. First part \( = \frac{1}{2} + \frac{3}{2}\ln 3 - 3\ln 2 \); second part \( = \frac{3}{2} - 3\ln 3 + 3\ln 2 \).
4. Sum \( = 2 - \frac{3}{2}\ln 3 = \frac{4 - 3\ln 3}{2} \).

Teacher's Note:
1. Sketch the tent-shaped graph of \( y = 2 - |2 - x| \) with its peak at \( (2, 2) \) and the hyperbola before integrating.
2. Reject intersection points that fall outside the piece of the modulus function you are using.

Answer: (a) \( \frac{1}{3} \)

Explanation:
1. Substitute \( x = \frac{y^2}{4} \) in the line: \( \frac{y^2}{2} - 3y + 4 = 0 \), so \( y^2 - 6y + 8 = 0 \), giving \( y = 2, 4 \) and the points \( (1, 2) \) and \( (4, 4) \).
2. Area \( = \int_1^4 \left(2\sqrt{x} - \frac{2x + 4}{3}\right)dx = \frac{28}{3} - 9 = \frac{1}{3} \).

Teacher's Note:
1. The parabola lies above the line between the two intersection points here, so subtract line from curve.
2. Evaluate \( \int 2\sqrt{x}\,dx \) and the linear part separately to avoid arithmetic slips.

Answer: (c) \( 8 \)

Explanation:
1. The curve consists of \( y = x^3 \) and \( y = -x^3 \), which are mirror images in the x-axis.
2. Area \( = 2\int_0^2 x^3\,dx = 2\cdot\frac{16}{4} = 8 \).

Teacher's Note:
1. A modulus on y gives a curve symmetric about the x-axis; compute one half and double.
2. Do not forget the factor 2, which is the commonest mistake here.

Answer: (d) \( \frac{1}{3} \) sq. units

Explanation:
1. The parabolas meet at \( (0, 1) \). The line \( y = \frac{1}{4} \) cuts \( y = (x - 1)^2 \) at \( x = \frac{1}{2} \) and \( y = (x + 1)^2 \) at \( x = -\frac{1}{2} \).
2. The bounded region is symmetric about the y-axis, so area \( = 2\int_0^{1/2}\left[(x - 1)^2 - \frac{1}{4}\right]dx \).
3. \( = 2\left[\frac{(x - 1)^3}{3} - \frac{x}{4}\right]_0^{1/2} = 2\left[\left(-\frac{1}{24} - \frac{1}{8}\right) + \frac{1}{3}\right] = \frac{1}{3} \).

Teacher's Note:
1. For a region between two curves with the y-axis as a mirror line, integrate over half the width and double.
2. Write down the intersection points before setting up the integral.

Answer: (a) \( \frac{\pi}{2} \)

Explanation:
1. \( \sqrt{y} \) needs \( y \ge 0 \). For \( x \ge 0 \) the conditions reduce to \( \sqrt{x} \le y \le \sqrt{2 - x^2} \). For \( x < 0 \), \( \sqrt{y} \ge -x \) gives \( y \ge x^2 \), and \( y^2 \ge x \) then holds automatically.
2. The two boundaries meet at \( (1, 1) \) and \( (-1, 1) \), so the region lies between \( x = -1 \) and \( x = 1 \).
3. Area \( = \int_{-1}^{1}\sqrt{2 - x^2}\,dx - \int_{-1}^{0}x^2\,dx - \int_0^1 \sqrt{x}\,dx \).
4. \( \int_{-1}^{1}\sqrt{2 - x^2}\,dx = 1 + \frac{\pi}{2} \), so area \( = 1 + \frac{\pi}{2} - \frac{1}{3} - \frac{2}{3} = \frac{\pi}{2} \).

Teacher's Note:
1. Translate each inequality into a boundary curve for each sign of x before drawing the region.
2. The integral of \( \sqrt{a^2 - x^2} \) uses the standard formula with an inverse sine term.

Answer: (a) \( \frac{3\pi a^2}{8} \)

Explanation:
1. Eliminating \( t \) gives the astroid \( x^{2/3} + y^{2/3} = a^{2/3} \), symmetric in all four quadrants.
2. Area \( = 4\int_0^{a} y\,dx \) with \( dx = -3a\cos^2 t \sin t\,dt \), which becomes \( 12a^2\int_0^{\pi/2}\sin^4 t\cos^2 t\,dt \).
3. \( \int_0^{\pi/2}\sin^4 t\cos^2 t\,dt = \frac{\pi}{32} \), so area \( = 12a^2 \cdot \frac{\pi}{32} = \frac{3\pi a^2}{8} \).

Teacher's Note:
1. Parametric curves are integrated as \( \int y\,\frac{dx}{dt}\,dt \) with the limits converted to values of \( t \).
2. Use the Wallis formula for \( \int_0^{\pi/2}\sin^m t\cos^n t\,dt \) instead of expanding.

Answer: (d) \( (-1, 0) \)

Explanation:
1. The square has area \( 16 \), so the region inside the parabola (between it and the line \( x = 2 \)) must have area \( 8 \).
2. Let the vertex be \( (-\alpha, 0) \). The parabola has the x-axis as its axis and the chord \( PS \) of length \( 4 \) at distance \( 2 + \alpha \) from the vertex.
3. Area of a parabolic segment \( = \frac{2}{3} \times \text{base} \times \text{height} = \frac{2}{3}\times 4 \times (2 + \alpha) = 8 \).
4. So \( 2 + \alpha = 3 \), \( \alpha = 1 \), and the vertex is \( (-1, 0) \).

Teacher's Note:
1. The area of a parabolic segment is two-thirds of the enclosing rectangle on the chord and its distance from the vertex.
2. Use a symmetric figure to turn a bisecting condition into a single equation.

Answer: (d) \( (6 - \pi) : \pi \)

Explanation:
1. On \( 0 \le x \le 1 \) the curves \( y = x^2 \) and \( y = \sin\frac{\pi x}{2} \) meet at \( x = 0 \) and \( x = 1 \), with the sine curve above.
2. Upper part \( = \int_0^1\left(\sin\frac{\pi x}{2} - x^2\right)dx = \frac{2}{\pi} - \frac{1}{3} = \frac{6 - \pi}{3\pi} \).
3. Lower part \( = \int_0^1 x^2\,dx = \frac{1}{3} \).
4. Ratio \( = \frac{6 - \pi}{3\pi} : \frac{1}{3} = (6 - \pi) : \pi \).

Teacher's Note:
1. Find the area under the upper curve and the area under the dividing curve separately, then take the difference.
2. Quote the ratio in the order the question uses: part above the dividing curve to the part below it.

Answer: (b) \( 2\pi \)

Explanation:
1. On \( \left(\frac{\pi}{2}, \pi\right] \): \( f(x) = 2 - f(\pi - x) = 2 - \sin(\pi - x) = 2 - \sin x \).
2. Area on \( [0, \pi] \) \( = \int_0^{\pi/2}\sin x\,dx + \int_{\pi/2}^{\pi}(2 - \sin x)\,dx = 1 + (\pi - 1) = \pi \).
3. Since \( f(x) = f(2\pi - x) \), the graph on \( [\pi, 2\pi] \) is a mirror image about \( x = \pi \), so it has the same area.
4. Total area \( = 2\pi \).

Teacher's Note:
1. Use the functional relations to write f explicitly on each interval before integrating.
2. A symmetry about a vertical line \( x = c \) means areas on the two sides are equal.

Answer: (a) \( \frac{2}{e} \)

Explanation:
1. Equating, \( x(e^x - e^{-x}) = 0 \) gives \( x = 0 \), so the curves meet at the origin; for \( 0 < x \le 1 \), \( xe^x > xe^{-x} \).
2. Area \( = \int_0^1 x(e^x - e^{-x})\,dx \). Integrate by parts: \( = \left[x(e^x + e^{-x})\right]_0^1 - \int_0^1 (e^x + e^{-x})\,dx \).
3. \( = \left(e + \frac{1}{e}\right) - \left(e - \frac{1}{e}\right) = \frac{2}{e} \).

Teacher's Note:
1. Integration by parts with \( u = x \) and \( dv = e^x - e^{-x} \) is the simplest route.
2. Check the sign: the upper curve for positive x is the one with \( e^x \).

Answer: (d) \( \frac{7}{12} \)

Explanation:
1. Both curves pass through \( (0, 0) \) and \( (1, 0) \) and, on \( (0, 1) \), \( 2x - 2x^2 > x\ln x \) (the first is positive, the second negative).
2. Area \( = \int_0^1 (2x - 2x^2 - x\ln x)\,dx \). Use \( \int_0^1 x\ln x\,dx = -\frac{1}{4} \).
3. Area \( = \left(1 - \frac{2}{3}\right) + \frac{1}{4} = \frac{7}{12} \).

Teacher's Note:
1. Remember \( \int_0^1 x\ln x\,dx = -\frac{1}{4} \); it appears often in area and series questions.
2. Test a point such as \( x = \frac{1}{2} \) to decide which curve is on top.

Answer: (b) \( 3 + \ln 2 \)

Explanation:
1. \( \max(|x|, |y|) \le 1 \) is the square \( -1 \le x, y \le 1 \), of area \( 4 \).
2. The part removed is where \( xy > \frac{1}{2} \) inside the square: in the first and third quadrants, each piece lies between \( y = \frac{1}{2x} \) and \( y = 1 \) for \( \frac{1}{2} \le x \le 1 \).
3. Removed area \( = 2\int_{1/2}^{1}\left(1 - \frac{1}{2x}\right)dx = 2\left[x - \frac{1}{2}\ln x\right]_{1/2}^{1} = 1 - \ln 2 \).
4. Required area \( = 4 - (1 - \ln 2) = 3 + \ln 2 \).

Teacher's Note:
1. Describe the region as a big shape minus the parts excluded by the second condition.
2. \( xy > \frac{1}{2} \) can only occur where x and y have the same sign.