- #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
SUMMARY
The discussion focuses on calculating the area under a curve using definite integrals. The integrals presented are $\displaystyle \int_{-1}^0 9 - 9^{-x} \, dx + \int_0^2 9 - 3^x \, dx$ and $\int_0^{\sqrt{\frac{\pi}{3}}} 2x sec^2(x^2)dx$. The second integral is transformed using the substitution $u = x^2$, leading to $\int_0^{\frac{\pi}{3}} sec^2(u)du$. The approach is confirmed to be correct, utilizing the chain rule for antiderivatives.
PREREQUISITES- Understanding of definite integrals
- Familiarity with the chain rule in calculus
- Knowledge of trigonometric functions, specifically secant
- Experience with integration techniques and substitutions
- Study the properties of definite integrals in calculus
- Learn about integration techniques involving trigonometric functions
- Explore substitution methods in integral calculus
- Practice solving integrals involving exponential functions
Students and professionals in mathematics, particularly those studying calculus, as well as educators looking to enhance their understanding of integration techniques.
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 $\int_0^{\frac{\pi}{3}} sec^2(u)du$.
Similar threads
High School
Question about units for "area under curve"
- · Replies 2 ·
- · Replies 2 ·
- · Replies 2 ·
- · Replies 5 ·
- · Replies 3 ·
- · Replies 2 ·
- · Replies 2 ·
High School
Explain Integration to me, please
- · Replies 5 ·
Undergrad
Area under curve using Excel question
- · Replies 2 ·
- · Replies 3 ·