Question
Using integration, Find the area bounded by the the triangle whose vartices are $(2, 1), (3, 4)$ and $(5, 2)$.
✓
Answer
Consider the points A(2, 1), B(3, 4) and C(5, 2)
We need to find area of shaded triangle ABC
Equation of AB is
$\text{y}-1=\Big(\frac{4-3}{3-2}\Big)(\text{x}-2)$
$\Rightarrow\text{x}-3\text{y}-5=0\ ...(\text{i})$
Equation of BC is
$\text{y}-4=\Big(\frac{2-4}{5-3}\Big)(\text{x}-3)$
$\Rightarrow\text{x}=\text{y}-7=0\ ...(\text{ii})$
Equation of CA is
$\text{y}-2=\Big(\frac{2-1}{5-2}\Big)(\text{x}-5)$
$\Rightarrow\text{x}-3\text{y}+2=0\ ...(\text{iii})$
Area of $\triangle\text{ABC}= \text{Area of}\ \triangle\text{ABC}+ \text{Area of }\triangle\text{ABC}$ in the,
Consided point $P(x_1, y_2)$ on AB and $Q(x_1, y_1)$ on AD
Thus, the area of appoximating with length = $|y_2 - y_1|$ from $x = 2$, to $x = 3$
$\therefore$ Area of $\triangle\text{ABC}=\int\limits_{2}^{3}|\text{y}_{2}-\text{y}_{1}|\text{dx} $
$\Rightarrow\text{A}=\int\limits_{2}^{3}(\text{y}_{2}-\text{y}_{1})\text{dx} $
$\Rightarrow\text{A}=\int\limits_{2}^{3}(3\text{x}-5)-\Big(\frac{\text{x}-1}{3}\Big)\text{dx}$
$\Rightarrow\text{A}=\int\limits_{2}^{3}\frac{(9\text{x}-15-\text{x}-1)}{3}\text{dx}$
$\Rightarrow\text{A}=\int\limits_{2}^{3}\frac{(8\text{x}-16)}{3}\text{dx}$
$\Rightarrow\text{A}=\frac{1}{3}\Big[8\text{x}^{2}-16\text{x}\Big]\text{dx}$
$\Rightarrow\text{A}=\frac{1}{3}(68-64)$
$\Rightarrow\text{A}=\frac{4}{3}\ \text{sq.}\ \text{units}$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Find the equation of the curve that passes through the point (0, a) and is such that at any point (x, y) on it, the product of its slope and the ordinate is equal to the abscissa.
Using differentials, find the approximate values of the following:
$\sqrt{49.5}$
→$\sqrt{49.5}$
If $l_i, m_i, n_i; i = 1, 2, 3$ denotes the direction cosines of three mutually perpendicular vectors in space, prove that $AA^T = I$, where
$\text{A}=\begin{bmatrix}\text{l}_1&\text{m}_1&\text{n}_1\\\text{l}_2&\text{m}_2&\text{n}_2\\\text{l}_3&\text{m}_3&\text{n}_3 \end{bmatrix}.$
→$\text{A}=\begin{bmatrix}\text{l}_1&\text{m}_1&\text{n}_1\\\text{l}_2&\text{m}_2&\text{n}_2\\\text{l}_3&\text{m}_3&\text{n}_3 \end{bmatrix}.$
Maximum Z = 2x + 3y
Subject to
$\text{x}+\text{y}\geq1$
$10\text{x}+\text{y}\geq5$
$\text{x}+10\text{y}\geq1$
$\text{x},\text{y}\geq0$
→Subject to
$\text{x}+\text{y}\geq1$
$10\text{x}+\text{y}\geq5$
$\text{x}+10\text{y}\geq1$
$\text{x},\text{y}\geq0$
If $\text{y}=\tan^{-1}\Big(\frac{\sqrt{1+\text{x}}-\sqrt{1-\text{x}}}{\sqrt{1+\text{x}}+\sqrt{1+\text{x}}}\Big),$ find $\frac{\text{dy}}{\text{dx}}.$
→Find the equation of the line passing through the points $\hat{\text{i}}+\hat{\text{j}}-3\hat{\text{k}}$ and perpendicular to the lines $\vec{\text{r}}=\hat{\text{i}}+\lambda\big(2\hat{\text{i}}+\hat{\text{j}}-3\hat{\text{k}}\big)$ and $\vec{\text{r}}=\big(2\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}\big)+\mu\big(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big).$
→Solve the following differential equation:
$\big(\cot^{-1}\text{y} + \text{x}\big)\text{dy}= \big(1 + \text{y}^2\big) \text{dx}$
→$\big(\cot^{-1}\text{y} + \text{x}\big)\text{dy}= \big(1 + \text{y}^2\big) \text{dx}$
A wire of length l is cut into two parts. One part is bent into a circle and the other into a square. Show that the sum of the areas of the circle and the square is the least if the radius of the circle is half of the side of the square.
Without expanding, show that the values of the following determinant are zero:
$\begin{vmatrix}\sin^2\text{A}&\cot\text{A}&1\\\sin^2\text{B}&\cot\text{B}&1\\\sin^2\text{C}&\cot\text{C}&1\end{vmatrix}$
→$\begin{vmatrix}\sin^2\text{A}&\cot\text{A}&1\\\sin^2\text{B}&\cot\text{B}&1\\\sin^2\text{C}&\cot\text{C}&1\end{vmatrix}$
Form the differential equation of the family of circle in the secound qudrant and touching the coordinate axes.