1. ⇒ (MHT CET 2023 11th May Evening Shift )

The slope of the tangent to a curve $y = f (x)$ at $(x, f (x))$ is $2 x + 1$ . If the curve passes through the point $(1, 2)$ , then the area (in sq. units), bounded by the curve, the $X$ -axis and the line $x = 1$ , is

A. $\frac{3}{2}$

B. $\frac{4}{3}$

C. $\frac{5}{6}$

D. $\frac{1}{12}$

Correct Option is (C)

MHT CET 2023 11th May Evening Shift Mathematics - Area Under The Curves Question 3 English Explanation

According to the given condition, $\frac{d y}{d x} = 2 x + 1$

Integrating w.r.t. $x$ , we get $y = x^{2} + x + c$

As curve passes through $(1, 2)$ , we get $2 = (1)^{2} + 1 + c \Rightarrow c = 0$

$∴$ The equation of the curve is $y = x^{2} + x$ .

$∴$ Required area

$\begin{aligned} = \int_{0}^{1} (x^{2} + x) d x \\ = {[\frac{x^{3}}{3} + \frac{x^{2}}{2}]}_{0}^{1} \\ = \frac{1}{3} + \frac{1}{2} \\ = \frac{5}{6} sq. units \end{aligned}$

2. ⇒ (MHT CET 2023 10th May Evening Shift )

The area bounded by the $X$ -axis and the curve $y = x (x - 2) (x + 1)$ is

A. $\frac{37}{12}$ sq. units

B. $\frac{27}{12}$ sq. units

C. $\frac{37}{4}$ sq. units

D. $\frac{27}{13}$ sq. units

Correct Option is (A)

$\begin{aligned} For X-axis, \\ y = 0 \\ ∴ x (x - 2) (x + 1) = 0 \\ \Rightarrow x = 0 or x = 2 or x = - 1 \end{aligned}$

MHT CET 2023 10th May Evening Shift Mathematics - Area Under The Curves Question 5 English Explanation

Required area $= \int_{- 1}^{0} y d x + | \int_{0}^{2} y d x |$

$\begin{aligned} = \int_{- 1}^{0} x (x - 2) (x + 1) d x + | \int_{0}^{2} x (x - 2) (x + 1) d x | \\ = \int_{- 1}^{0} (x^{3} - x^{2} - 2 x) d x + | \int_{0}^{2} (x^{3} - x^{2} - 2 x) d x | \\ = {[\frac{x^{4}}{4} - \frac{x^{3}}{3} - x^{2}]}_{- 1}^{0} + | {[\frac{x^{4}}{4} - \frac{x^{3}}{3} - x^{2}]}_{0}^{2} | \\ = \frac{5}{12} + | - \frac{8}{3} | \\ = \frac{37}{12} sq. units \end{aligned}$

3. ⇒ (MHT CET 2023 10th May Morning Shift )

Area of the region bounded by the curve $y = \sqrt{49 - x^{2}}$ and $X$ -axis is

A. $49 π$ sq. units

B. $\frac{49 π}{2}$ sq. units

C. $\frac{49 π}{4}$ sq. units

D. $98 π$ sq. units

Correct Option is (B)

MHT CET 2023 10th May Morning Shift Mathematics - Area Under The Curves Question 6 English Explanation

$\begin{aligned} Required area & = 2 \int_{0}^{7} \sqrt{49 - x^{2}} d x \\ = 2 {[\frac{x}{2} \sqrt{49 - x^{2}} + \frac{49}{2} \sin^{- 1} (\frac{x}{7})]}_{0}^{7} \\ = 2 [\frac{7}{2} \sqrt{49 - 49} + \frac{49}{2} \sin^{- 1} (\frac{7}{7})] - 0 \\ = 2 \times \frac{49}{2} \times \frac{π}{2} \\ = \frac{49 π}{2} sq. units \end{aligned}$

4. ⇒ (MHT CET 2023 9th May Evening Shift )

The area (in sq. units) bounded by the curve $y = x | x |, X$ -axis and the lines $x = - 1$ and $x = 1$ is

A. $\frac{2}{3}$

B. $\frac{1}{3}$

C. 1

D. $\frac{4}{3}$

Correct Option is (A)

$\begin{aligned} y = x | x | \dots [Given] \\ Required area \\ = \int_{- 1}^{1} x | x | d x \\ = 2 \int_{0}^{1} x^{2} d x ... [∵ Area is always positive] \\ = 2 \times {[\frac{x^{3}}{3}]}_{0}^{1} \\ = 2 \times (\frac{1}{3} - 0) = \frac{2}{3} sq.units \end{aligned}$

5. ⇒ (MHT CET 2021 21th September Evening Shift )

The area bounded by the parabola $y^{2} = x$ , the straight line $y = 4$ and $Y$ axis is

A. $2 \sqrt{7}$ sq. units

B. $\frac{64}{3}$ sq. units

C. $\frac{16}{3}$ sq. units

D. $7 \sqrt{2}$ sq. units

Correct Option is (B)

MHT CET 2021 21th September Evening Shift Mathematics - Area Under The Curves Question 10 English Explanation

Refer figure

Required area is shaded.

Point of intersection of $y^{2} = x$ and $y = 4$ is $A \equiv (16, 4)$ .

$\begin{aligned} ∴ A & = \int_{0}^{4} y^{2} d y \\ = {[\frac{y^{3}}{3}]}_{0}^{4} = \frac{64}{3} sq. units \end{aligned}$

⇒Application of Definite Integration