Area Under The Curve - JEE Advanced Previous Year Questions with Solutions

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Q. Area of the region bounded by the curve $y=e^{x}$ and $\operatorname{lines} x=0$ and $y=e$ is

(A) e – 1

(B) $\int_{1}^{\mathrm{e}} l \mathrm{n}(\mathrm{e}+1-\mathrm{y}) \mathrm{d} \mathrm{y}$

(D) $\int_{1}^{\mathrm{e}} \ln \mathrm{y} \mathrm{d} \mathrm{y}$

[JEE 2009, 4]

Sol. (B,C,D)

$\mathrm{A}=\int_{1}^{e} \ln y \mathrm{d} y$

Apply $=\int_{1}^{e} \ln (e+1-y) d y$

$\mathrm{Also}, \mathrm{A}=\operatorname{ar}(\mathrm{O} \mathrm{ABC})-\operatorname{ar}(\mathrm{OABD})$

$=\mathrm{e}-\int_{1}^{e} \mathrm{e}^{\mathrm{x}} \mathrm{d} \mathrm{x}$

Q. Comprehension (3 questions together) :

Consider the poly nomial $\mathrm{f}(\mathrm{x})=1+2 \mathrm{x}+3 \mathrm{x}^{2}+4 \mathrm{x}^{3} .$ Let $\mathrm{s}$ be the sum of all distinct real roots

of $\mathrm{f}(\mathrm{x})$ and let $\mathrm{t}=|\mathrm{s}| .$

(i) The real number s lies in the interval

(ii) The area bounded by the curve y = f(x) and the lines x = 0, y = 0 and x = t, lies in the interval

(iii) The function f'(x) is

(A) increasing in $\left(-\mathrm{t},-\frac{1}{4}\right)$ and decreasing in $\left(-\frac{1}{4}, \mathrm{t}\right)$

(B) decreasing in $\left(-\mathrm{t},-\frac{1}{4}\right)$ and increasing in $\left(-\frac{1}{4}, \mathrm{t}\right)$

(D) decreasing in $(-\mathrm{t}, \mathrm{t})$

[JEE 2010, 3+3+3]

Sol. ((i) $\mathrm{C} \quad$ (ii) $\mathrm{A} \quad$ (iii) $\mathrm{B}$)

Q. (A) Let the straight line $\mathrm{x}=\mathrm{b}$ divide the area enclosed by $\mathrm{y}=(1-\mathrm{x})^{2}, \mathrm{y}=0$ and $\mathrm{x}=0$ into two

parts $\mathrm{R}_{1}(0 \leq \mathrm{x} \leq \mathrm{b}) \mathrm{d}$ and $\mathrm{R}_{2}(\mathrm{b} \leq \mathrm{x} \leq 1)$ such that $\mathrm{R}_{1}-\mathrm{R}_{2}=\frac{1}{4} .$ Then $\mathrm{b}$ equals

(A) $\frac{3}{4}$

(B) $\frac{1}{2}$

(D) $\frac{1}{4}$

(B) Let $f:[-1,2] \rightarrow[0, \infty)$ be a continuous function such that $f(\mathrm{x})=f(1-\mathrm{x})$ for all $\mathrm{x} \in[-1,2]$. Let $\mathrm{R}_{1}=\int_{-1}^{2} \mathrm{x} f(\mathrm{x}) \mathrm{d} \mathrm{x}$, and $\mathrm{R}_{2}$ be the area of the region bounded by $\mathrm{y}=f(\mathrm{x}), \mathrm{x}=-1, \mathrm{x}=2$,and the x-axis. Then-

(A) $\mathrm{R}_{1}=2 \mathrm{R}_{2}$

(B) $\mathrm{R}_{1}=3 \mathrm{R}_{2}$

(D) $3 \mathrm{R}_{1}=\mathrm{R}_{2}$

[JEE 2011, 3+3]

Sol. ((A) B (B) C )

Q. The area enclosed by the curve $y=\sin x+\cos x$ and $y=|\cos x-\sin x|$ over the interval $\left[0, \frac{\pi}{2}\right]$ is

(A) $4(\sqrt{2}-1)$

(B) $2 \sqrt{2}(\sqrt{2}-1)$

(D) $2 \sqrt{2}(\sqrt{2}+1)$

[JEE(Advanced) 2013, 2M]

Sol. (B)

Q. Let $\mathrm{F}(\mathrm{x})$$=\int_{x}^{x^{2}+\frac{\pi}{6}}$ $2 \cos ^{2}$ tdt for all $\mathrm{x} \in \mathbb{U}$ and $f:\left[0, \frac{1}{2}\right] \rightarrow[0, \infty)$ be a continuous function. For $\mathrm{a} \in\left[0, \frac{1}{2}\right],$ if $\mathrm{F}^{\prime}(\mathrm{a})+2$ is the area of the region bounded by x = 0, y = 0, y = ƒ(x) and x = a, then ƒ(0) is

[JEE 2015, 4M, –0M]

Sol. 3

Q. Area of the region $\{ \left(\mathrm{x}, \mathrm{y}) \in \mathbb{D}^{2}: \mathrm{y} \geq \sqrt{|\mathrm{x}+3|, 5 \mathrm{y} \leq \mathrm{x}+9 \leq 15\}}$ is equal to – \right.

(A) $\frac{1}{6}$

(B) $\frac{4}{3}$

(D) $\frac{5}{3}$

[JEE(Advanced) 2016]

Sol. (C)

Q. If the line $x=\alpha$ divides the area of region $R=\left\{(x, y) \in \square^{2}: x^{3} \leq y \leq x, 0 \leq x \leq 1\right\}$ into two equal parts, then

(A) $\frac{1}{2}<\alpha<1$

(B) $\alpha^{4}+4 \alpha^{2}-1=0$

(D) $2 \alpha^{4}-4 \alpha^{2}+1=0$

[JEE(Advanced) 2017]

Sol. (A,D)

Q. Let $f:[0, \infty) \rightarrow \square$ be a continuous function such that $f(\mathrm{x})=1-2 \mathrm{x}+\int_{0}^{\mathrm{x}} \mathrm{e}^{\mathrm{x}-\mathrm{t}} f(\mathrm{t}) \mathrm{d} \mathrm{t}$ for all $\mathrm{x} \in[0, \infty) .$ Then, which of the following statement(s) is (are) TRUE?

(A) The curve y = ƒ(x) passes through the point (1, 2)

(B) The curve y = ƒ(x) passes through the point (2, –1)

(D) The area of the region $\{(x, y) \in[0,1] \times \square: f(x) \leq y \leq \sqrt{1-x^{2}}\}$ is $\frac{\pi-1}{4}$

[JEE(Advanced) 2018]

Sol. (B,C)

Q. A farmer $F_{1}$ has a land in the shape of a triangle with vertices at $P(0,0), Q(1,1)$ and $R(2,)$ 0 . From this land, a neighbouring farmer $F_{2}$ takes away the region which lies between the side $P Q$ and a curve of the form $y=x^{n}(n>1) .$ If the area of the region taken away by the farmer $F_{2}$ is exactly $30 \%$ of the area of $\triangle P Q R,$ then the value of $n$ is

Sol. 4

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