NCERT Exemplar Class 12 Maths Integrals Solutions - Class 12 M...

Strategic angle. Switching to a horizontal-strip integral (dy instead of dx) avoids the square root entirely, because both boundaries become polynomials in y. The line y=3x becomes x=y/3, the parabola y²=9x becomes x=y²/9, and integration turns into a clean polynomial evaluation.

1. Rewrite both boundaries as functions of y: y²=9x⇒ x=y²9 (parabola), y=3x⇒ x=y3 (line).
2. Identify the y-range. At the origin both curves give y=0; at the upper intersection (1,3) both give y=3. So y runs from 0 to 3 and the strip width is the x-difference between the line and the parabola.
3. For fixed y∈[0,3], the left boundary is the parabola and the right boundary is the line. Check at y=1: parabola gives x=1/9≈ 0.11, line gives x=1/3≈ 0.33, so the line is to the right of the parabola.
4. Set up the integral as right minus left: A=₀³(y3-y²9)dy.
5. Antidifferentiate term by term: ∫y3 dy=y²6, ∫y²9 dy=y³27.
6. Evaluate from 0 to 3: [y²6-y³27]₀³ =96-2727=32-1=12.

Why this matters. Choosing dx vs dy is a routine optimisation in board problems: if one direction gives polynomial limits and the other gives square roots, pick the polynomial one. Two independent methods giving the same answer is a strong cross-check.

A=12 sq units.

Picture-first. The two parabolas are reflections of each other in the line y=x, so by symmetry the area can also be found as twice the area between y=√2px and the line y=x from x=0 to x=2p.

1. By symmetry about y=x, the region is split into two congruent halves by the line y=x. Call the lower half R₁ (between the line y=x and the parabola x²=2py) and the upper half R₂ (between y²=2px and y=x). Then A=2(R₂).
2. Compute Area(R₂)=₀^2p(√2px-x) dx: aligned &=[2√2p3x^3/2-x²2]₀^2p
   &=8p²3-(2p)²2=8p²3-2p²=2p²3. aligned
3. Double: A=2·2p²3=4p²3.

Why this matters. Spotting the y=x symmetry shortens the algebra; many JEE problems with reflected parabolas, ellipses, or circles yield to the same trick.

A=4p²3 sq units.

Quick reading. The cubic factorisation gives a single real intersection at x=2, and the line x=0 cuts the cubic at the origin and the line y=x+6 at (0,6), so the region is a closed curvilinear "triangle" with vertices at (0,0), (0,6), (2,8).

1. Confirm the bounds: at x=0 the line is at y=6 and the cubic is at y=0, so the strip width at x=0 is 6. At x=2 both curves meet at y=8, so the strip width is 0. The region is bounded above by y=x+6, below by y=x³, left by x=0, and closes at x=2.
2. Apply the area formula: A=₀²((x+6)-x³)dx =[x²2+6x-x⁴4]₀².
3. Substitute: A=2+12-4=10 sq units.

Why this matters. Spotting one rational root of a cubic by inspection (try ± 1,± 2,± 3,± 6 for x³-x-6=0) is a standard exam technique that saves time.

A=10 sq units.

Strategic angle. Use the symmetry about y=x: the line bisects the enclosed lens, so the area equals twice the area between y=2√x and the line y=x on [0,4].

1. By symmetry, A=2₀⁴(2√x-x) dx.
2. Integrate: ∫ (2√x-x) dx = 43x^3/2-x²2. Evaluate from 0 to 4: 43· 8 - 162=323-8 =32-243=83.
3. Double: A=2·83=163.

Why this matters. The y=x symmetry is a common JEE pattern (e.g. y²=x and x²=y); recognising it cuts the computation in half.

A=163 sq units.

Quick reading. Same template as Q1 with the slope changed from 3 to 1; the line is now flatter, so the intersection moves further out (from x=1 to x=9) and the enclosed area is much larger.

1. Use a dy integral instead. y=x becomes x=y and y²=9x becomes x=y²9.
2. For y∈[0,9], the line x=y is to the right of the parabola x=y²/9 (check y=3: x=3 vs x=1).
3. Compute A=₀⁹(y-y²9)dy =[y²2-y³27]₀⁹ =812-72927=812-27. Now 812-27=81-542=272.

Why this matters. The dy approach avoids the square root, trading it for a clean quadratic. Both routes give the same answer, which is a good cross-check.

A=272 sq units.

Strategic angle. A cleaner evaluation uses the difference of polynomials and the identity _a^b(b-x)(x-a) dx=(b-a)³6 for parabolic segments.

1. Write the integrand as -(x²-x-2)=-(x-2)(x+1)=(2-x)(x+1).
2. Apply the standard formula _-1²(2-x)(x+1) dx=(2-(-1))³6=276=92.

Why this matters. Recognising the parabolic-segment formula 16(b-a)³ when both factors are linear collapses the integration to one line. Very useful for MCQ time-pressure.

A=92 sq units.

Picture-first. Use horizontal strips: for each y, the strip runs from x=y²8 (parabola) to x=2 (line), and y ranges from -4 to 4 (where y²=8· 2 gives y=± 4).

1. Set up A=_-4⁴(2-y²8)dy =2₀⁴(2-y²8)dy using even symmetry of the integrand.
2. Compute ₀⁴(2-y²8)dy =[2y-y³24]₀⁴ =8-6424=8-83=163.
3. Double: A=323.

Why this matters. For parabolas of the form y²=4ax cut off by a vertical line x=h, the dy method yields a clean polynomial. Try both methods and pick the one with fewer surds.

A=323 sq units.

Strategic angle. The square root √a²-x² begs for the trig substitution x=asinθ, because then a²-x²=a²(1-sin²θ)=a²cos²θ and the surd disappears. Here a=2, so we try x=2sinθ.

1. Set x=2sinθ, so dx=2cosθ dθ, and √4-x²=√4-4sin²θ=2|cosθ|=2cosθ (positive in θ∈[-π/2,π/2]).
2. Update the limits. When x=-2: sinθ=-1, so θ=-π/2. When x=2: sinθ=1, so θ=π/2. The interval x∈[-2,2] maps to θ∈[-π/2,π/2] bijectively.
3. Substitute everything: A=_-π/2^π/22cosθ· 2cosθ dθ =4_-π/2^π/2cos²θ dθ.
4. Use cos²θ=1+cos 2θ2 on the integrand to obtain 4_-π/2^π/21+cos 2θ2 dθ =2_-π/2^π/2(1+cos 2θ) dθ.
5. Integrate: ∫(1+cos 2θ) dθ=θ+sin 2θ2.
6. Evaluate: 2[θ+sin 2θ2]_-π/2^π/2 =2[(π2+0)-(-π2+0)] =2(π)=2π. (sin(π)=0, so the second term vanishes at both limits.)

Why this matters. The x=asinθ substitution is the go-to move whenever √a²-x² appears in a definite integral. It always pairs with the half-angle identity to reduce the integral to a sum of constants and sinusoids; memorise this two-step pattern.

A=2π sq units.

Quick reading. Standard area-under-curve drill: power rule with fractional exponent.

1. Rewrite 2√x=2x^1/2.
2. Apply ∫ xⁿ dx=xⁿ⁺¹n+1 with n=12: ∫ 2x^1/2 dx=2x^3/23/2=4x^3/23.
3. Plug in x=1 and x=0: 43-0=43.

Why this matters. The power rule with fractional exponents shows up everywhere; keep √x=x^1/2 in mind so you can integrate without thinking.

A=43 sq units.

Picture-first. A straight-line boundary above the x-axis with vertical sides at x=2 and x=8 traces a trapezium. Either integrate or apply the trapezium area formula.

1. Compute heights: y(2)=5(2)+72=172 and y(8)=5(8)+72=472.
2. Width w=8-2=6.
3. Trapezium area =12(b₁+b₂)h =12(172+472)· 6 =12·642· 6=16· 6=96.

Why this matters. Whenever the integrand is linear, the "area under a line" is a trapezium; using elementary geometry for the sanity check catches arithmetic errors fast.

A=96 sq units.

Strategic angle. Recognise that y=√x-1 is the curve y=√x shifted right by 1 unit, so its area between x=1 and x=5 equals the area of y=√x between x=0 and x=4, no substitution needed.

1. Translate: the area under y=√x-1 on [1,5] equals the area under y=√x on [0,4].
2. Compute the resulting integral directly: ₀⁴√x dx=[23x^3/2]₀⁴ =23· 8=163.

Why this matters. Translations f(x-h) shift the graph right by h without changing areas; this saves time and avoids substitution mistakes.

A=163 sq units.

Quick reading. Use the trig substitution x=asinθ, turning the surd into acosθ.

1. Let x=asinθ, dx=acosθ dθ. When x=0, θ=0; when x=a, θ=π/2. √a²-x²=acosθ (positive in [0,π/2]).
2. Substitute: A=₀^π/2acosθ· acosθ dθ =a²₀^π/2cos²θ dθ.
3. Use cos²θ=1+cos 2θ2: A=a²2₀^π/2(1+cos 2θ) dθ =a²2[θ+sin 2θ2]₀^π/2 =a²2·π2=π a²4.

Why this matters. The trig substitution is a fundamental move; learning it on this clean example pays dividends in every ellipse/circle problem.

A=π a²4 sq units.

Structural observation. Switch to dy: the curves become x=y² (parabola) and x=y (line); on [0,1], y≥ y², so the line is to the right of the parabola.

1. Set up A=₀¹(y-y²) dy=[y²2-y³3]₀¹.
2. Evaluate: 12-13=3-26=16.

Why this matters. Symmetry across y=x shows the parabola y²=x and its reflection y=x² also enclose 1/3 of area (twice 1/6); a useful family of standard results.

A=16 sq units.

Quick reading. The integrand -x²+x+2 factors as -(x-2)(x+1), so it equals (2-x)(x+1). Apply the parabolic-segment formula.

1. Rewrite: -(x-2)(x+1)=(2-x)(x+1).
2. Use _a^b(b-x)(x-a) dx=(b-a)³6 with a=-1, b=2: A=(2-(-1))³6=276=92.

Why this matters. The "16(b-a)³" formula appears whenever you integrate a quadratic between its two roots a and b; recognising it saves the polynomial evaluation.

A=92 sq units.

Strategic angle. Cross-check by splitting the region into two x-integrals at x=3 (where the line x=2y+3 crosses the x-axis): one piece from x=0 to x=3 under the curve and above the x-axis, and a second piece from x=3 to x=9 between the curve and the line. The sum should also equal 9.

1. Identify the dividing abscissa x=3 where the line meets the x-axis. For x<3 the bottom of the region is the x-axis; for x>3 the bottom is the line y=x-32.
2. Piece 1 (from x=0 to x=3): ₀³√x dx =[23x^3/2]₀³ =23(3)^3/2=23· 3√3=2√3.
3. Piece 2 (from x=3 to x=9): the top is y=√x and the bottom is y=x-32. Strip width √x-x-32. So ₃⁹(√x-x-32)dx =[23x^3/2-(x-3)²4]₃⁹.
4. Evaluate at x=9: 23· 9^3/2-(9-3)²4 =23· 27-364=18-9=9. At x=3: 23· 3√3-0=2√3. Difference: 9-2√3.
5. Add the two pieces: 2√3+(9-2√3)=9. The irrational terms cancel exactly, confirming the answer.

Why this matters. Two independent setups (one dy, one split-dx) giving the same answer is a strong confirmation. The dy-route is shorter, but seeing both reinforces the geometric picture and catches silly arithmetic errors.

A=9 sq units.

Picture-first. The parabola cuts the circle into two pieces; the region we want is the "moon"-shaped sliver outside the parabola but inside the circle (or its mirror image). By symmetry we compute the upper half.

1. Half-disc area (right half of the circle, x≥ 2) plus the strip 0≤ x≤ 2 above the parabola = total upper half of the bounded region.
2. Total area of upper half of circle: 12π r²=12π(4)=2π.
3. Subtract the upper half of the parabolic segment inside the circle: the parabola's upper branch from (0,0) to (2,2) bounds the region we DON'T want above the x-axis. Area under parabola from x=0 to x=2 (above x-axis): ₀²√2x dx=83 (computed above).
4. So upper half of bounded region = 2π-83 minus the area of the upper-left rectangular piece of the disc that the parabola does not cover. Actually it's simpler: the region we want (in y≥ 0) is bounded above by the upper semicircle, below by the upper parabola, on 0≤ x≤ 2. Area = (upper half of disc, the part with 0≤ x≤ 2) - (parabola's region 0≤ x≤ 2, 0≤ y≤√2x). Half-disc has area 2π, of which the left half (x≤ 2) is π. So upper half of bounded region = π-83.
5. By x-axis symmetry, total bounded area = 2(π-83)=2π-163.

Why this matters. Decomposing the region using known disc sectors (half-disc, quarter-disc) replaces hard integrals with simple multiplications.

A=2π-163 sq units.

Quick reading. By periodicity, the half-period [0,π] contributes area 2; the next half [π,2π] contributes the same 2 (just below the axis). Total 4.

1. Area under one arch: ₀^πsin x dx=[-cos x]₀^π=2.
2. Two arches between 0 and 2π, each of area 2.
3. Total: 2· 2=4.

Why this matters. Sine has period 2π and the area under each arch is a fixed constant (2 for unit amplitude). For sin nx between 0 and 2π, the area is 2n·1|n|·|n|=4 (there are 2|n| arches, each of area 2|n|).

A=4 sq units.

Structural observation. Integrate with respect to y instead. The left edge of the triangle is the segment AB from A(-1,1) to B(0,5), while the right side consists of two pieces: segment AC from A to C(3,2) on the lower part, and segment BC from B to C on the upper part. The transition occurs at y=2 (the y-coordinate of C). So we set up two horizontal-strip integrals split at y=2.

1. Left boundary: AB in inverse form, y=4x+5⇒ x=y-54 on y∈[1,5].
2. Right boundary changes at y=2 (the y-coordinate of C). For y∈[1,2]: right side is AC, i.e. x=4y-5 (inverse of y=x+54). For y∈[2,5]: right side is BC, i.e. x=5-y.
3. Compute: ₁²((4y-5)-y-54)dy =₁²16y-20-y+54dy=₁²15y-154dy. =154[y²2-y]₁² =154((2-2)-(12-1))=154·12=158.
4. And: ₂⁵((5-y)-y-54)dy =₂⁵20-4y-y+54dy=₂⁵25-5y4dy. =54[5y-y²2]₂⁵ =54((25-252)-(10-2)) =54(252-8)=54·92=458.
5. Total: 158+458=608=152.

Why this matters. Two integration routes confirming the same geometric area is the kind of cross-check graders love to see in long-answer questions.

A=152 sq units.

Picture-first. The disc has area 16π a². The parabola y²=6ax cuts a "tongue" out of the disc; the bounded region is the intersection. Split the upper half (y≥ 0) into a parabolic cap (under the parabola, for 0≤ x≤ 2a) and a circular cap (under the upper semicircle, for 2a≤ x≤ 4a), and double for the lower half.

1. Identify the split. The parabola and circle intersect at x=2a (in the upper half, y=2√3 a). For 0≤ x≤ 2a the parabola is the binding constraint (y≤√6ax); for 2a≤ x≤ 4a the circle is the binding constraint (y≤√16a²-x²).
2. Parabolic piece. Compute ₀^2a√6ax dx=√6a·23(2a)^3/2 =4a3√12a²=8√3 a²3. (This is the area below the upper parabola, above the x-axis, on [0,2a].)
3. Circular piece, computed geometrically. The chord x=2a is at distance 2a from the disc's centre (0,0); the chord's half-length above the axis is 2√3 a. The half-angle subtended at centre is sin^-1(2√3 a/(4a))=sin^-1(√3/2)=π3. Half-sector area above the axis (sector of angle π/3): 12R²·π3=12· 16a²·π3=8π a²3. Subtract the right-angled triangle of base 2a (from x=0 to x=2a along the chord-foot) and height 2√3 a: 12· 2a· 2√3 a=2√3 a². Net circular cap (upper, between x=2a and x=4a): 8π a²3-2√3 a².
4. Sum the upper-half pieces: 8√3 a²3+8π a²3-2√3 a² =8√3 a²-6√3 a²3+8π a²3 =2√3 a²+8π a²3.
5. Double for the full region (by x-axis symmetry): A=4√3 a²+16π a²3=4a²3(4π+√3).

Why this matters. Decomposing into a parabolic cap plus a circular segment (sector minus triangle) lets you bypass the √R²-x² integral entirely. The half-angle sin^-1(√3/2)=π/3 is a special-angle that's worth spotting on sight.

A=4a²3(4π+√3) sq units.

Strategic angle. Once the three vertices are known, 12|det| delivers the area in one line; integration just verifies. Use the shoelace-style determinant formula based on the two edge vectors emanating from a chosen vertex.

1. Recall the vertices found in the main solution: A(0,1), B(2,3), C(4,-1).
2. Form the two edge vectors from A: AB⃗=(2-0,3-1)=(2,2) and AC⃗=(4-0,-1-1)=(4,-2).
3. Apply the cross-product (signed-area) determinant: Δ=12|detpmatrix2 & 2 4 & -2pmatrix| =12|2·(-2)-2· 4|=12|-4-8|=12· 12=6.
4. Compare with the integration answer: 6. The two routes agree.
5. Sanity check via the symmetric formula: Δ=12|x₁(y₂-y₃)+x₂(y₃-y₁)+x₃(y₁-y₂)| =12|0(3+1)+2(-1-1)+4(1-3)| =12|0-4-8|=6. Same.

Why this matters. The integration route shows the splitting mechanism; the determinant route gives the same answer instantly. Long-answer questions usually require both: integration in the body, geometry as a sanity check at the end. Knowing two routes is insurance against arithmetic slip-ups.

A=6 sq units.

Quick reading. The three slopes are 4, -1, and 14. The product of slopes of L₁ and L₃ is 4·14=1≠ -1, so they are not perpendicular, but they do bound a triangle together with L₂. Notice that the slopes 4 and 14 are reciprocals, hinting at a reflective symmetry of the configuration about the line y=x+const.

1. Confirm vertices from pairwise intersection (done in the main solution): A(-1,1), B(0,5), C(3,2).
2. Edge vectors from A: AB⃗=(0-(-1),5-1)=(1,4), AC⃗=(3-(-1),2-1)=(4,1).
3. Cross-product determinant: Δ=12|1· 1-4· 4|=12|1-16|=152.
4. Verify via the symmetric formula: Δ=12|(-1)(5-2)+0(2-1)+3(1-5)| =12|-3+0-12|=152. Both give 152.
5. Connect to Q18: this is the same triangle stated by its three side-equations rather than its three vertices, so the answer must agree.

Why this matters. Recognising a problem you have already solved in a different presentation is a high-value skill in exam settings. Always pause and check: have I seen these vertices before?

A=152 sq units.

Picture-first. Each "lobe" of y=2cos x has the same area because amplitude and arch length are constant. Compute one lobe (0 to π) and multiply.

1. Area of one half-arch of |2cos x| from 0 to π/2: ₀^π/22cos x dx=2.
2. By symmetry, each quarter-period [π/2 wide] of |2cos x| has the same area 2.
3. On [0,2π] there are exactly 4 such quarter-periods (from 0 to π/2, π/2 to π, π to 3π/2, 3π/2 to 2π). Total: 4· 2=8.

Why this matters. Once you realise the area in one quarter-period is 2, the full period [0,2π] gives 8 instantly for any amplitude; more generally, |Acos x| on a full period gives total area 4|A|.

A=8 sq units.

Structural observation. The function y=1+|x+1| is piecewise linear with a corner at x=-1. So the region between y=0 and the curve, for x∈[-3,3], is the union of two trapeziums that share the vertical edge at x=-1. We avoid integration entirely and use the trapezium-area formula 12(sum of parallel sides).

1. Compute heights at the three relevant x-values: y(-3)=1+|-2|=3, y(-1)=1+0=1, y(3)=1+4=5.
2. Left trapezium: vertices (-3,0),(-3,3),(-1,1),(-1,0), parallel sides y(-3)=3 (left) and y(-1)=1 (right), width -1-(-3)=2. Area =12(3+1)· 2=4· 22=4.
3. Right trapezium: vertices (-1,0),(-1,1),(3,5),(3,0), parallel sides y(-1)=1 (left) and y(3)=5 (right), width 3-(-1)=4. Area =12(1+5)· 4=6· 42=12.
4. Total: 4+12=16. Matches the integration answer.

Why this matters. For any piecewise-linear function above the x-axis, splitting at the corners and computing trapezium areas by hand is faster and less error-prone than mechanical integration. The same idea generalises: a polygon under any piecewise-linear curve decomposes into trapeziums.

A=16 sq units.

Quick reading. The integrand is cos x-sin x, whose antiderivative is sin x+cos x. At x=π/4 both sin(π/4) and cos(π/4) equal 1/√2, so the sum is √2. At x=0 the sum is sin 0+cos 0=0+1=1. Difference gives the answer immediately.

1. Set up the integral as ₀^π/4(cos x-sin x) dx (cos is above sin on [0,π/4], the y-axis is the left boundary, and the intersection at x=π/4 closes the region).
2. Antidifferentiate: ∫(cos x-sin x) dx=sin x+cos x.
3. Recall the special-angle values sin(π/4)=cos(π/4)=1√2=√22, so sin(π/4)+cos(π/4)=2√2=√2.
4. Evaluate at the lower limit: sin 0+cos 0=0+1=1.
5. Subtract: √2-1. Match: option (C).

Why this matters. Recognising special-angle values is faster than re-deriving them; for MCQs on [0,π/2], expect π/4, π/6, π/3 to appear, with values 1/√2, 1/2, √3/2.

Option (C).

Quick reading. Same intersections as Q6 (roots of x²-x-2=0), so the polynomial integral equals 92. Here we just have the extra factor 14 from the equations y=x²4 and y=x+24.

1. Reuse _-1²((x+2)-x²) dx=92 (parabolic segment formula with b-a=3 gives 276).
2. Multiply by the 14 scaling: 14·92=98.

Why this matters. Spotting structural similarity to a previous problem saves entire integration steps.

Option (D).

Picture-first. Do not integrate; just identify the curve as the upper half of a circle and apply 12π r².

1. Square both sides of y=√16-x² (valid for y≥ 0): y²=16-x², equivalent to x²+y²=16. This is a circle of radius r=4 centred at the origin.
2. Since y≥ 0, only the upper half of this circle is the curve in question.
3. Half-disc area: 12π r²=12·π· 4² =16π2=8π.
4. Compare with the four options. Only (A) matches. Distractor analysis: (B) 20π would correspond to an ellipse with a· b=20, not a circle; (C) 16π is the FULL disc area, not the half; (D) 256π comes from mistakenly using r²=16²=256 instead of r=4.

Why this matters. Geometric recognition reduces a two-minute integration to a five-second check; under exam time pressure, this is decisive. Always check whether a "curve" is secretly a piece of a standard shape (circle, ellipse, line) before reaching for the integral.

Option (A).

Structural observation. In the first quadrant, the region bounded by the x-axis (below), the line y=x (above), and the arc of the circle x²+y²=32 is a circular sector subtending angle π4 (i.e. 45^∘) at the centre. The sector's area is a fraction π/42π=18 of the full disc.

1. Identify the disc radius: r²=32⇒ r=4√2, full disc area =π r²=32π.
2. The line y=x makes angle π/4 with the positive x-axis. The region inside the disc, in the first quadrant, bounded below by the x-axis and above by y=x, is a circular sector of angle π/4.
3. Fraction of full disc: π/42π=18.
4. Sector area: 18· 32π=4π.
5. Match option (B).
6. Cross-check via the formula A=12r²θ: A=12· 32·π4=32π8=4π. Same.

Why this matters. For MCQ-style geometry, fraction-of-disc arguments beat integration every time. Whenever the bounding line is y=x, y=√3 x, or another standard-angle line, the sector formula gives the answer in two lines.

Option (B).

Quick reading. Each half-arch of |cos x| has area 1 (for unit amplitude). On [0,π] there are two such half-arches, total 2.

1. Each lobe of |cos x|: area ₀^π/2cos x dx=1.
2. Two lobes on [0,π] (the half above the axis on [0,π/2] and the half below on [π/2,π]): total 2.

Why this matters. Memorise: each half-arch of unit-amplitude sine or cosine has area 1.

Option (A).

Strategic angle. Apply the parabolic-segment formula on the y-axis: _a^b(b-y)(y-a) dy=(b-a)³6.

1. Write 2y-y²=-(y²-2y)=-y(y-2)=y(2-y), which equals (2-y)(y-0) in standard form with a=0, b=2.
2. Apply the formula: (2-0)³6=86=43.

Why this matters. The 16(b-a)³ shortcut is the fastest route for any parabola-meets-line area, but you must factor the integrand as (b-y)(y-a) with the roots showing.

Option (A).

Quick reading. Half-arch of unit-amplitude sine on [0,π/2] has area 1.

1. Standard result: ₀^π/2sin x dx=1.
2. Match: option (D).

Why this matters. Internalise the four "unit area" benchmarks: ₀^π/2sin x = 1, ₀^πsin x = 2, ₀^2π|sin x| = 4, and the analogous cosine integrals.

Option (D).

Picture-first. The ellipse is a circle of radius 1 stretched horizontally by 5 and vertically by 4, so its area scales by 5· 4=20 relative to the unit disc's area π.

1. Unit disc has area π.
2. Stretch by a=5 horizontally: area becomes 5π.
3. Stretch by b=4 vertically: area becomes 5· 4π=20π.

Why this matters. The "linear-scaling" argument for ellipse area extends to any affine transformation: area scales by |det J| where J is the Jacobian.

Option (A).

Quick reading. Unit circle, area π.

1. Disc area =π r²=π· 1²=π.

Why this matters. The unit-circle area π is the foundational constant of plane geometry; commit it deeply.

Option (B).

Picture-first. A trapezium with vertical sides at x=2 (height 3) and x=3 (height 4), and a slanted top from (2,3) to (3,4).

1. Heights: y(2)=3, y(3)=4.
2. Width: w=3-2=1.
3. Area =12(3+4)· 1=72.

Why this matters. For any linear y=f(x) above the x-axis on [a,b], the area is 12(f(a)+f(b))(b-a) by the trapezium rule; this is exact for linear functions.

Option (A).

Quick reading. The line x=2y+3, with y∈[-1,1] and the y-axis as the implied left boundary, encloses a trapezium lying on its side. The two parallel sides are horizontal segments of length x(-1)=1 and x(1)=5; the width (vertical extent) is 1-(-1)=2. Apply the trapezium-area formula.

1. Evaluate the endpoints: x(-1)=2(-1)+3=1 and x(1)=2(1)+3=5. Both are positive, so the trapezium lies entirely to the right of the y-axis.
2. Identify the parallel sides (the two horizontal segments at y=± 1) and the width (the vertical distance between them): parallel sides =1 and 5; width =2.
3. Apply the formula A=12(b₁+b₂)· w=12(1+5)· 2=12· 12=6.
4. Match: option (C).
5. Verify via integration: _-1¹(2y+3) dy =[y²+3y]_-1¹=(1+3)-(1-3)=4-(-2)=6. Same.

Why this matters. The trapezium-area formula works just as well when the strip is horizontal; swap the roles of x and y in your head and the formula carries over verbatim. For linear x=f(y) between two horizontal lines, this is exact and faster than antidifferentiation.

Option (C).