Using integration find the area of the triangular region whose sides have the equations y =2x+1,y=3x+1 and x=4.
Using integration find the area of the triangular region whose sides have the equations y =2x+1,y=3x+1 and x=4.
Solution and Explanation
The equations of side of the triangle are y=2x+1,y=3x+1, and x=4.
To solving these equations, we obtain the vertices of triangle as A(0,1), B(4,13), and C
(4,9).
It can be observe that,
Area(ΔACB)=Area(OLBAO)-Area(OLCAO) =
\[\int_{0}^{4} (3x+1) \,dx\]\[-\int_{0}^{4} (2x+1) \,dx\]
=[\(\frac{3x^2}{2}\)+x]40-[\(\frac{2x^2}{2}\)+x]40
=(24+4)-(16+4)
=28-20
=8 units.
Top Questions on applications of integrals
- Let \(\sum_{n=0}^{\infty }\frac{n^{3}((2n)!+(2n-1)(n!))}{(n!)((2n)!)}=ae+\frac{b}{e}+c\), where $a, b, c \in Z$ and $e=\displaystyle\sum_{n=0}^{\infty} \frac{1}{n !}$ Then $a^2-b+c$ is equal to _______
- JEE Main - 2023
- Mathematics
- applications of integrals
- $\displaystyle\lim _{x \rightarrow \infty} \frac{(\sqrt{3 x+1}+\sqrt{3 x-1})^6+(\sqrt{3 x+1}-\sqrt{3 x-1})^6}{\left(x+\sqrt{x^2-1}\right)^6+\left(x-\sqrt{x^2-1}\right)^6} x^3$
- JEE Main - 2023
- Mathematics
- applications of integrals
The area of the region given by \(\left\{(x, y): x y \leq 8,1 \leq y \leq x^2\right\}\) is :
- JEE Main - 2023
- Mathematics
- applications of integrals
- Find area bounded by the curves \(y\) = \(max\{sin x, cos x\}\) and \(x-axis\) between \(x = - \pi \)and \(x = \pi\)
- JEE Main - 2023
- Mathematics
- applications of integrals
- \(I(x) = \int \frac{x^2(xsec^2x + tanx)}{(xtanx+1)^2} dx\). If \(I(0)=1\) then \(I(\frac{\pi}{4})\) is equal to
- JEE Main - 2023
- Mathematics
- applications of integrals
Questions Asked in CBSE CLASS XII exam
- Find the inverse of each of the matrices, if it exists. \(\begin{bmatrix} 1 & 3\\ 2 & 7\end{bmatrix}\)
- For what values of x,\(\begin{bmatrix} 1 & 2 & 1 \end{bmatrix}\)\(\begin{bmatrix} 1 & 2 & 0\\ 2 & 0 & 1 \\1&0&2 \end{bmatrix}\)\(\begin{bmatrix} 0 \\2\\x\end{bmatrix}\)=O?
- Find the inverse of each of the matrices,if it exists \(\begin{bmatrix} 2 & 1 \\ 7 & 4 \end{bmatrix}\)
What is the Planning Process?
- CBSE CLASS XII - 2023
- Planning process steps
- Find the inverse of each of the matrices,if it exists. \(\begin{bmatrix} 2 & 3\\ 5 & 7 \end{bmatrix}\)
Notes on applications of integrals
- Application of Integrals Mathematics
- NCERT Solutions For Class 12 Mathematics Chapter 8 Applications of the Integrals
- Trapezoidal Rule Mathematics
- Properties of Definite Integrals Mathematics
- NCERT Solutions for Applications of Integrals Exercise 8 1 Solutions Mathematics
- NCERT Solutions for Applications of Integrals Exercise 8 2 Solutions Mathematics
Concepts Used:
Area between Two Curves
Integral calculus is the method that can be used to calculate the area between two curves that fall in between two intersecting curves. Similarly, we can use integration to find the area under two curves where we know the equation of two curves and their intersection points. In the given image, we have two functions f(x) and g(x) where we need to find the area between these two curves given in the shaded portion.
Area Between Two Curves With Respect to Y is
If f(y) and g(y) are continuous on [c, d] and g(y) < f(y) for all y in [c, d], then,
