Question

solve two question and find answer

Answer 1

Area Under the Curve mathematically means that we have to find the definite integral with the given limits.


We asked to find the area under the curve $y=5e^x$, the X-Axis, and between $x=0$ and $x=2$


In other words, we have to find the Integral of $5e^x$ with limits as $x=0$ to $x=2$

[A graph is attached for visual representation]

Let the Area be $A$


Then we have:


$\displaystyle A = \int\limits_0^2 5e^x \, dx \\ \\ \\ \implies A=5\left[\, e^x\, \right]_0^2 \\ \\ \\ \implies A=5[e^2-e^0]\\ \\ \\ \implies \boxed{\bold{A=5(e^2-1) \, \, sq. \, units}}$


Thus, The Answer is Option (C).


__________________________________



In the second question, we are given that a car starts from rest, and moves with constant acceleration.


Let us suppose the constant acceleration $a$.


Let distance covered in n seconds be denoted by $s_n$


Then, we have:


$\displaystyle s=ut+\frac{1}{2}at^2 \\ \\ \\ \implies s_n =(0)(n)+\frac{1}{2}a(n)^2 \ \\ \\ \implies \boxed{s_n = \frac{an^2}{2}}$



Now, We need the ratio of distance travelled in $n^{th}$ second (say $s_{n^{th}}$) to the distance travelled in $n$ seconds ($s_n$)


To get the distance travelled in $n^{th}$ second, we will use this logic:


Distance travelled in the nth second is equal to the distance travelled in (n-1) seconds subtracted from the distance travelled in n seconds.


That is,


$s_{n^{th}}=s_n-s_{n-1}$


For example, suppose we want to find the distance travelled in the 4th second. Then we first find the distance travelled in complete 4 seconds, and from it, we subtract the distance travelled in 3 seconds.


Now, we can get the answer.


$\displaystyle Ans = \frac{s_{n^{th}}}{s_n} \\ \\ \\ \implies \frac{s_{n^{th}}}{s_n} = \frac{s_n-s_{n-1}}{s_n} \\ \\ \\ \implies \frac{s_{n^{th}}}{s_n} = \frac{\frac{an^2}{2}-\frac{a(n-1)^2}{2}}{\frac{an^2}{2}}\\ \\ \\ \implies \frac{s_{n^{th}}}{s_n} = \frac{n^2-(n-1)^2}{n^2} \\ \\ \\ \left[\text{Using the Identity }a^2-b^2=(a+b)(a-b)\right] \\ \\ \\ \implies \frac{s_{n^{th}}}{s_n} = \frac{(n+n-1)(n-n+1)}{n^2} \\ \\ \\ \implies \frac{s_{n^{th}}}{s_n} = \frac{(2n-1)(1)}{n^2}$


$\displaystyle \implies \frac{s_{n^{th}}}{s_n} = \frac{2n-1}{n^2} \\ \\ \\ \implies \frac{s_{n^{th}}}{s_n} = \frac{2n}{n^2}-\frac{1}{n^2} \\ \\ \\ \implies \boxed{\bold{\frac{s_{n^{th}}}{s_n}=\frac{2}{n}-\frac{1}{n^2}}}$



Thus, The answer is Option (C)