Find the area bounded by the parabola `x^(2) = y`, x-axis and the line

1.	Find the area bounded by the parabola `x^(2) = y`, x-axis and the line `y = 1`
Answer» Graph of ` y = x^(2)` Area OEBD = Area OAEO `= underset(0)overset(1)int\|x\| dy = underset(0)overset(1)intsqrt(y) dy = 2/3`

Discussion

Consider the integral `I=int_(0)^(2pi)(dx)/(5-2cosx)` Making the substitution `"tan"1/2x=t`, we have `I=int_(0)^(2pi)(dx)/(5-2cosx)=int_(0)^(0)(2dt)/((1+t^(2))[5-2(1-t^(2))//(1+t^(2))])=0` The result is obviously wrong, since the integrand is positive and consequently the integral of this function cannot be equal to zero. Find the mistake.
The area bounded by the curves `y=cosx` and `y=sinx` between the ordinates `x=0` and `x=(3pi)/2` isA. `4sqrt(2) + 2`B. `4sqrt(2) +1`C. `4sqrt(2) + 1`D. `4sqrt(2) - 2`
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If `I_n=int_0^pi e^x(sinx)^n dx ,` then `(I_3)/(I_1)` is equal toA. `3//5`B. `1//5`C. `1`D. `2//5`
Prove that:`I_n=int_0^oox^(2n+1)e^-x^2dx=(n !)/2,n in Ndot`
If f(x) is continuous and `int_(0)^(9)f(x)dx=4`, then the value of the integral `int_(0)^(3)x.f(x^(2))dx` isA. 2B. 18C. 16D. 4
`lim_(trarr0) int_(0)^(2pi)(|sin(x+t)-sinx|)/(|t|)dx` equalsA. 2B. 4C. 43469D. 1
Given that `lim_(nrarroo) sum_(r=1)^(n) (log_(e)(n^(2)+r^(2))-2log_(e)n)/n = log_(e)2+pi/2-2`, then evaluate : ` lim_(nrarroo) (1)/(n^(2m))[(n^(2)+1^(2))^(m)(n^(2)+2^(2))^(m) "......"(2n^(2))^(m)]^(1//n)`.
`lim_(nrarroo) sum_(r=0)^(n-1) (1)/(sqrt(n^(2)-r^(2)))`
`lim_(nrarroo) (((n+1)(n+2)"........"3n)/(n^(2n)))^(1//n)` is equal to :A. `(27)/(e^(2))`B. `(9)/(e^(2))`C. `3 log3-2`D. `(18)/(e^(4))`