Class 12

Math

Calculus

Application of Integrals

Find the area of the parabola $y_{2}=4ax$bounded by its latus rectum.

Connecting you to a tutor in 60 seconds.

Get answers to your doubts.

Consider two curves $C_{1}:y_{2}=4[y ]xandC_{2}:x_{2}=4[x ]y,$ where [.] denotes the greatest integer function. Then the area of region enclosed by these two curves within the square formed by the lines $x=1,y=1,x=4,y=4$ is $38 squ˙nits$ (b) $310 squ˙nits$ $311 squ˙nits$ (d) $411 squ˙nits$

Using integration, find the area of the region enclosed between the two circles $x_{2}+y_{2}=4$ and $(x−2)_{2}+y_{2}=4.$

If $_{′}a_{prime}(a>0)$ is the value of parameter for each of which the area of the figure bounded by the straight line $y=1+a_{4}a_{2}−ax $ and the parabola $y=1+a_{4}x_{2}+2ax+3a_{2} $ is the greatest, then the value of $a_{4}$ is___

Find the area enclosed by the circle $x_{2}+y_{2}=a_{2}$.

Using the method of integration find the area bounded by the curve $∣x∣+∣y∣=1$.[Hint: The required region is bounded by lines $x+y=1,x−y=1,−x+y=1$and$−x−y=1]˙$

Find the area bounded by the curve $y=(x−1)(x−2)(x−3)$ lying between the ordinates $x=0andx=3.$

Find the area of the region bounded by the line $y=3x+2$, the x-axis and the ordinates $x=1andx=1$.