CBSE Class 12 (Science) Application of Integrals — practice questions
95 free MCQs with worked solutions. Tap any question for the answer + explanation, or practice them all in the app.
Practice CBSE Class 12 (Science) Application of Integrals in the app →Formulae of elementary geometry cannot find areas enclosed by:The area under a curve is built from a large number of thin:For a vertical strip of height y and width dx, dA equals:The area of one thin strip is called the:The total area comes from adding up all the:The definite integral is evaluated using which theorem?Strips used to find an area may be vertical or:If a curve lies below the x-axis, its integral comes out:When the integral is negative, the area is taken as its:If part of a curve is above the axis and part below, the areas are:For the region below the axis A1 is negative while A2 is:For a horizontal strip, the elementary area would be:Areas are found here for lines and arcs of circles, plus:The width of an elementary vertical strip is written as:The definite integral was first defined as the limit of:The height of a vertical elementary strip is given by:An elementary area is located at a position specified by some value of:Only the numerical value of an area is taken into consideration, so signs are:Finding the area of a circle here uses which curves in standard form?The area under y equals f of x is bounded by the curve, the x-axis and two:The definite integral was earlier calculated as the limit of a:The mathematician pictured alongside the introduction is:Cauchy lived from 1789 until:Besides areas under curves, the topic finds areas between lines and arcs of circles, parabolas and:The curves treated are taken in their forms that are:Definite integrals are evaluated using the Fundamental Theorem of:The area under a curve is pictured as composed of many thin strips that are:An elementary strip of height y and width dx has area dA equal to:In the strip formula, the height y is given by:If the curve lies below the x-axis, the computed area comes out:When the computed area is negative, the value taken is its:For areas, only the value that is taken into consideration is the:For a region in the first quadrant, y is taken as:Finding the area bounded by a curve is described as a way that is easy and:A later section treats the area between how many curves?The application of integrals studied here is finding:The strips used to build up area have width written as:The ordinates bounding the area are drawn at x equals a and:Curves whose areas are found include circles, parabolas and:A large number of very thin strips are added to approximate the:The definite integral's sign for a curve below the axis is:How large an area does the circle x² + y² = 49 bound? Use π = 22/7The area of the ellipse x²/16 + y²/9 = 1 equalsA plot lies under y = x² and above the x-axis between the ordinates 0 and 3. Its area isA ramp follows y = 2x + 1. How much area lies under it from x = 0 to 2?One full arch of the sine curve on [0, π] and the x-axis enclose an area ofOver a full period 0 to 2π, the cosine curve and the x-axis bound a total area ofBetween the ordinates x = 0 and x = 1, the exponential curve y = eˣ bounds an area ofThe hyperbola y = 1/x, the x-axis, x = 1 and x = e cut off a region of areaDetermine the area trapped by the root curve y = √x, the x-axis and the line x = 4The cubic y = x³ rises over [0, 2]. How large is the region beneath it?The line x = 3 cuts a cap from the parabola y² = 4x. Find the area of that capHow much area does y² = 4ax enclose with its focal chord perpendicular to the axis?A line of slope 1 through the origin meets y² = 4x again. Find the area between themFind the area shared by the line y = x and the parabola y = x² between their meeting pointsA horizontal line y = 4 caps the parabola y = x². Find the enclosed areaParabolas y² = x and x² = y meet at the origin and (1, 1). Find the area they encloseThe parabola y = x² and the chord y = 2x meet at (0, 0) and (2, 4). Find the area betweenThe triangle enclosed by y = |x| and the line y = 2 has areaFind the area between the V-shaped curve y = |x - 1| and the x-axis for -2 ≤ x ≤ 4Find the area of the triangle cut from the first quadrant by the line x + y = 2A line with intercepts a and b on the axes forms a triangle with them. Its area isArea in the first quadrant bounded by x² + y² = 4, x = 0 and x = 2 isWhat is the area of the semicircle given by y = √(25 - x²)?Area of the ellipse x²/a² + y²/b² = 1 lying in the first quadrant isTo find the area bounded by x = g(y), the y-axis and y = c, y = d, we use strips that areIf a curve lies below the x-axis from x = a to x = b, the area taken isThe integral of sin x over 0 to 2π is 0, but the area between the curve and the x-axis isHow big is the region closed in by the parabola y = x² - 4 and the x-axis?The inverted parabola y = 9 - x² and the x-axis enclose an area ofTaking all area as positive, the region between y = 3x + 2, the x-axis, x = -1 and x = 1 has areaFind the region between the line y = 2x, the x-axis and the ordinate x = 3Over [1, 2], the reciprocal-square curve y = 1/x² and the x-axis bound an area ofThe bell-shaped curve y = 1/(1 + x²) over [0, 1] bounds an area ofThe curve y = ln x, the x-axis and the line x = e enclose an area ofArea enclosed between y = x and y = x³ in the first quadrant isSine and cosine curves enclose a sliver between x = 0 and x = π/4. Its area isComplete the square in x² + y² = 2x and find the area it boundsArea of the region enclosed by 4x² + y² = 36 isThe parabola x² = 4y and the line y = x enclose what area?Integrating the hypotenuse line, find the area of the triangle with corners (0,0), (4,0), (4,3)Corners of a triangular plot are (1, 0), (2, 2) and (3, 1). How much area does it cover?The region between the parabola 3x² and the x-axis for x in [1, 2] has what area?Using horizontal strips, find the area bounded by y² = 4x, the y-axis and y = 3One hump of y = sin 2x on [0, π/2] encloses what area?A reflector section is y² = 16x cut off by x = 4. Find its areaFind the area of the triangle bounded by y = x, the x-axis and x + y = 4The area under y = √(a² - x²) for 0 ≤ x ≤ a isA circle with centre (2, 0) and equation (x - 2)² + y² = 9 bounds an area ofThe V-shaped curve y = |x| over [-3, 3] and the x-axis enclose how much area?Elementary geometry formulae are inadequate for finding areas that areThe area under y = f(x) from x = a to x = b is obtained by addingWhat is the area under the parabola y = x² over the symmetric interval [-2, 2]?The smaller part of x² + y² = a² cut off by the line x = a/√2 has areaArea between the x-axis and the parabola y = 4x - x² is