Practice: Finding The Area Under A Curve

Notice that $x^2+a$ is an even function, and we are finding the area on a symmetric range. Therefore, we can solve for $a$ by setting the area under the curve from $0$ to $3$ as $18$ . $\int_{-3}^3 x^2+a \text{ dx}=2\int_0^3 x^2+a \text{ dx}=\dfrac{x^3}{3}+ax]^3_0$ Plugging in $3$ , this becomes a simple linear equation to solve for $a$ . $\dfrac{x^3}{3}+ax]^3_0=\dfrac{27}{3}+3a=9+3a=18\Rightarrow a=\boxed{3}$

Now, we all know that area under a curve is integration of that curve's function... So, 36 = integral [x^{2} - a ] within the intervals -3 and 3..... we get \frac{x^{3}}{3} + 3a within the limits -3 and 3...... so, 36=18 + 6a........ therefore, \boxed{a=6}

As we know that the area of a curve= (integration of y.dx). since area of given curve=36(also given),. on simplifying the given integral by putting limit(3 & -3). we get a=3.

integral uper limit is 3 and lower limit is -3 f(x)= x^2+a=36 after integration is 6a=18 a=3

The area under the curve from 3 to -3 is the integral of (x^2+a)dx from 3 to -3. integ(3,-3)(x^2)=18. The total area is 36, so the remainng 18 is contributed by integ(3,-3)(a) which is equal to 6a. 6a=18 or a=3.

We know that the function $f(x)=x^2 +a$ is even because $f(3)=f(-3)=9+a$ . Because of this, $\int_{0}^{3} \! f(x) \, \mathrm{d}x=\int_{-3}^{0} \! f(x) \, \mathrm{d}x$ , and we can then get $\int_{-3}^{3}\! f(x) \, \mathrm{d}x=2\int_{0}^{3}\! f(x)\, \mathrm{d}x$ . Because $a$ is a constant it will raise the function $x^2$ on the y-axis, meaning that it will change the area under the curve by $a(3--3)$ . We then get $\int_{-3}^{3}\! x^2+a\, \mathrm{d}x=\int_{-3}^{3}\! x^2\, \mathrm {d}x+a(3--3)=18+6a=36, 6a=18, a=\boxed{3}$ .

afta integrating x^2+a w.r.t x; we have: (x^3)/3+ax , substituting limits from -3 to 3 for x; we get: 18+6a=36, solving for a.... we get a=3.

Solving By Integrals We Get 6a=18......Which Gives a=3

This problem is a simple matter of setting up an integral from -3 to 3, and working from there. That being done, it comes out looking something like (1/3)x^3 +3ax ] from (-3, 3). Since the integral is solved by substituting top value minus the bottom value for all the X's, you then set it equal to 36 and end up with the equation: 9 + 3a + 9 + 3a = 36, which becomes 18 + 6a = 36. This is probably the most difficult part, as algebra is mind bogglingly difficult, but after a few hours I managed to simplify it to 6a = 18, and ultimately get a = 3.

2 (x^3+3xa)/3 Alisacado no intervalo de 0 a 3, é: 2 (3^3+3 3 a=3*36 logo a=3