Application of Integrals — JEE MCQs with answers

Answer: A — The Remark says that if the area is negative we take its absolute value, and it prints $\left|\int_a^b f(x)\,dx\right|$. This is option A. Option B is the integral itself, which is negative here. Option C puts a minus sign in front of the absolute value and is again negative. Option D uses only the two end values of $f$ and involves no integration.

Answer: B — The Note says on pages 298-299: 'In this sense, the method of exhaustion can be regarded as an early method of integration.' So option B. Options A and C are terms the Note attaches to the work of Newton (the inverse method of tangents; the theory of fluxions), and option D echoes what it credits to the works of A.L. Cauchy; none of them is said of the method of exhaustion.

Answer: A — $\int_{-2}^{2}x^{3}\,dx=\left[\dfrac{x^{4}}{4}\right]_{-2}^{2}=\dfrac{16}{4}-\dfrac{16}{4}=0$. The integral keeps the sign of each part, and the part from $x=-2$ to $x=0$, which is $-4$, cancels the part from $x=0$ to $x=2$, which is $4$. The answer is option A. Option B ($8$) is the area of the region bounded by the curve, the x-axis and the ordinates $x=-2$ and $x=2$; that is not what is asked. Option C ($4$) is the integral from $0$ to $2$ only. Option D ($-4$) is the integral from $-2$ to $0$ only.

Answer: B — The first sentence of Section 8.1 on page 292 names 'triangles, rectangles, trapezias and circles'. Parallelograms are not in that list, so option B is the answer. Options A, C and D are three of the four printed figures (rectangles being the remaining one); 'trapezias' is the spelling printed in the book.

Answer: D — Putting $(1,2)$ into $y=kx^{2}$ gives $k=2$, so the curve is $y=2x^{2}$. The area is $\int_{0}^{1}2x^{2}\,dx=\left[\dfrac{2x^{3}}{3}\right]_{0}^{1}=\dfrac{2}{3}$, which is option D. Option A gives the area of the rectangle with sides $1$ and $2$, which is $2$. Option B gives the area of the region between the parabola and the y-axis up to $y=2$, which is $\int_{0}^{2}\sqrt{\dfrac{y}{2}}\,dy=\dfrac{4}{3}$. Option C joins the origin to $(1,2)$ by a straight line and takes the triangle below it, $\dfrac{1}{2}\times 1\times 2=1$.

Answer: C — Example 1 (page 294) finds the area enclosed by a circle, Example 2 (page 295) the area enclosed by an ellipse, Example 3 (stated on page 296) the area bounded by a line, the $x$-axis and two ordinates, and Example 4 (page 297) the area bounded by the cosine curve. No numbered Example takes a parabola, so option C. The word 'parabolas' is printed only in the scope sentence of Section 8.1; in the chapter a parabola is met only in exercise questions, for instance the curve $y^2 = 4x$ of Exercise 8.1. Options A, B and D are the curves of Examples 3, 1 and 2.

Answer: A — The solution says: as the region AOBA lies in the first quadrant, $y$ is taken as positive. So the integrand is $y=\sqrt{a^2-x^2}$ and the area is $4\int_0^a\sqrt{a^2-x^2}\,dx$. This is option A. Option D has the right sign but not the printed reason; the sign of $a$ does not decide the sign of $y$. Option C forgets that only the first-quadrant piece is integrated, the other three pieces being covered by the factor $4$. Option B picks the lower half of the circle, where the region AOBA does not lie.

Answer: B — Page 292 prints 'area between lines and arcs of circles, parabolas and ellipses (standard forms only)'. So option B. The hyperbola is not named anywhere in Section 8.1; options A, C and D each put it in place of one of the three printed curves (the ellipse, the parabola and the circle respectively).

Answer: D — The curve is $y=4+x$ for $x<0$ and $y=4-x$ for $x\ge 0$. It meets the x-axis at $x=-4$ and $x=4$, both inside the interval: it is below the axis from $-6$ to $-4$, above from $-4$ to $4$, and below from $4$ to $5$. Area $=\left|\int_{-6}^{-4}(4+x)\,dx\right|+\int_{-4}^{0}(4+x)\,dx+\int_{0}^{4}(4-x)\,dx+\left|\int_{4}^{5}(4-x)\,dx\right|=2+8+8+\frac{1}{2}=\frac{37}{2}$, which is option D. Option A is the value of the integral, $-2+16-\frac{1}{2}$, with the parts below the axis subtracted. Option B counts only the part above the axis. Option C splits only at $x=4$ and misses the crossing at $x=-4$: $\left|\int_{-6}^{4}(4-|x|)\,dx\right|+\frac{1}{2}=14+\frac{1}{2}$.

Answer: A — The first paragraph of Section 8.1 on page 292 ends: 'For that we shall need some concepts of Integral Calculus.' So option A. Option B names the other branch of calculus, which this sentence does not mention. Option C is the very tool the paragraph has just called inadequate for areas enclosed by curves. Option D is a phrase from the Historical Note (the name Newton gave his work), not from Section 8.1.