Roisincleary
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How do I do questions like this? (see picture)
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Definite integral with one limit given
- Thread starter Roisincleary
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Hello, and welcome to FMH! 
Let's examine the first one:
[MATH]I=\int_0^{\alpha} x^4+x^2\,dx[/MATH]
Now, using the FTOC, we may state:
[MATH]I=\left[\frac{x^5}{5}+\frac{x^3}{3}\right]_0^{\alpha}=\frac{\alpha^3}{15}\left(3\alpha^2+5\right)[/MATH]
This is equivalent to choice (b). Can you now try the second problem?
Let's examine the first one:
[MATH]I=\int_0^{\alpha} x^4+x^2\,dx[/MATH]
Now, using the FTOC, we may state:
[MATH]I=\left[\frac{x^5}{5}+\frac{x^3}{3}\right]_0^{\alpha}=\frac{\alpha^3}{15}\left(3\alpha^2+5\right)[/MATH]
This is equivalent to choice (b). Can you now try the second problem?
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To follow up:
2.) [MATH]I=\int_0^{\alpha} x^{\frac{1}{3}}-x^{-\frac{1}{3}}\,dx=\left[\frac{3}{4}x^{\frac{4}{3}}-\frac{3}{2}x^{\frac{2}{3}}\right]_0^{\alpha}=\frac{3\alpha^{\frac{2}{3}}}{4}\left(\alpha^{\frac{2}{3}}-2\right)[/MATH]
This is equivalent to choice (a).
3.) [MATH]I=\int_1^{\alpha}\frac{1}{4x-1}\,dx=\left[\frac{1}{4}\ln(4x-1)\right]_1^{\alpha}=\frac{1}{4}\ln\left(\frac{4\alpha-1}{3}\right)[/MATH]
This is equivalent to choice (d).
2.) [MATH]I=\int_0^{\alpha} x^{\frac{1}{3}}-x^{-\frac{1}{3}}\,dx=\left[\frac{3}{4}x^{\frac{4}{3}}-\frac{3}{2}x^{\frac{2}{3}}\right]_0^{\alpha}=\frac{3\alpha^{\frac{2}{3}}}{4}\left(\alpha^{\frac{2}{3}}-2\right)[/MATH]
This is equivalent to choice (a).
3.) [MATH]I=\int_1^{\alpha}\frac{1}{4x-1}\,dx=\left[\frac{1}{4}\ln(4x-1)\right]_1^{\alpha}=\frac{1}{4}\ln\left(\frac{4\alpha-1}{3}\right)[/MATH]
This is equivalent to choice (d).