- #1
jaychay
- 58
- 0
Can you please check it for me that I have done it wrong or not ?
Thank you in advance.
MHB Calculating Area Under a Curve: Is My Approach Correct?
Click For Summary
The discussion focuses on verifying the correctness of the approach to calculate the area under a curve using integrals. The first integral is correctly set up as the difference between the top and bottom functions, specifically $\int_{-1}^0 9 - 9^{-x} \, dx + \int_0^2 9 - 3^x \, dx$. The second problem involves the integral $\int_0^{\sqrt{\frac{\pi}{3}}} 2x sec^2(x^2)dx$, which is transformed using the substitution $u = x^2$. This substitution simplifies the integral to $\int_0^{\frac{\pi}{3}} sec^2(u)du$, confirming the setup is accurate. Overall, the calculations and transformations presented are valid for determining the area under the specified curves.
Can you please check it for me that I have done it wrong or not ?
Thank you in advance.
- #2
skeeter
- 1,103
- 1
w/ respect to x, top function - bottom function
$\displaystyle \int_{-1}^0 9 - 9^{-x} \, dx + \int_0^2 9 - 3^x \, dx$
integral w/respect to y is ok and it can be simplified ...
$\displaystyle \dfrac{3}{2\ln{3}} \int_1^9 \ln{y} \, dy$
secant function integral is set up correctly ... antiderivative is rather easy to see if you recognize the chain rule
$\displaystyle \int_{-1}^0 9 - 9^{-x} \, dx + \int_0^2 9 - 3^x \, dx$
integral w/respect to y is ok and it can be simplified ...
$\displaystyle \dfrac{3}{2\ln{3}} \int_1^9 \ln{y} \, dy$
secant function integral is set up correctly ... antiderivative is rather easy to see if you recognize the chain rule
- #3
HOI
- 921
- 2
The second problem looks pretty standard- take x from 0 to $\sqrt{\frac{\pi}{3}}$ and, for each x, y from 0 to $2xsec^2(x^2)$.
The area is $\int_0^{\sqrt{\frac{\pi}{3}}} 2x sec^2(x^2)dx$.
Let $u= x^2$ so that $du= 2xdx$. When x= 0, u=0 and when x= $\sqrt{\frac{\pi}{3}}$, $u= \frac{\pi}{3}$.
The integral becomes $\int_0^{\frac{\pi}{3}} sec^2(u)du$.
The area is $\int_0^{\sqrt{\frac{\pi}{3}}} 2x sec^2(x^2)dx$.
Let $u= x^2$ so that $du= 2xdx$. When x= 0, u=0 and when x= $\sqrt{\frac{\pi}{3}}$, $u= \frac{\pi}{3}$.
The integral becomes $\int_0^{\frac{\pi}{3}} sec^2(u)du$.
Hello!
I tried to calculate projections of curl using rotation of coordinate system but I encountered with following problem.
Given:
##rot_xA=\frac{\partial A_z}{\partial y}-\frac{\partial A_y}{\partial z}=0##
##rot_yA=\frac{\partial A_x}{\partial z}-\frac{\partial A_z}{\partial x}=1##
##rot_zA=\frac{\partial A_y}{\partial x}-\frac{\partial A_x}{\partial y}=0##
I rotated ##yz##-plane of this coordinate system by an angle ##45## degrees about ##x##-axis and used rotation matrix to...
Similar threads
- · Replies 2 ·
- · Replies 2 ·
- · Replies 2 ·
- · Replies 5 ·
- · Replies 2 ·
- · Replies 2 ·
- · Replies 5 ·
- · Replies 2 ·
- · Replies 3 ·