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$$ \int_0^1 \frac{1 + 3x +5x^3}{\sqrt{x}}\ dx$$

My idea for this was to break each numerator into its own fraction as follows

$$ \int_0^1 \left(\frac{1}{\sqrt{x}} + \frac{3x}{\sqrt{x}} + \frac{5x^3}{\sqrt{x}}\right)dx$$

$$ \int_0^1 (x^{-1/2} + 3x^{1/2} +5x^{5/2})\ dx $$

$$ \int_0^1 2x^{1/2} + 2x^{3/2} + \frac{10}{7}x^{7/2} $$

Not really sure where to go from there. Should I sub 1 in for the x values and let that be the answer?

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    $\begingroup$ You seem to have miswritten the last step. $\int_0^1 (x^{-1/2} + 3x^{1/2} +5x^{5/2}) dx = 2x^{1/2} + 2x^{3/2} + \frac{10}{7}x^{7/2}|_{x=0}^1$ $\endgroup$ – WaveX Mar 24 '17 at 16:13
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It looks like you've already done the integration correctly, but forgot to take off the integration sign. The last line should be

$$\left[2x^{1/2} + 2x^{3/2} + \frac{10}{7}x^{7/2} \right]_0^1$$

which you can evaluate by substituting $x=1$ and $x=0$, and subtracting.

You asked whether you should substitute $x=1$: this will happen to give you the right answer, but only coincidentally, because the expression becomes $0$ when $x=0$.

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Hint: the formula $$\int x^c\, dx = \frac{x^{c+1}}{c+1} + C$$ applies to all real numbers $c \neq -1$, not just integers.

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  • $\begingroup$ Ok.. Not sure what you are saying? Should I apply this formula to my final step? Or perhaps to the formula before I integrate it ? $\endgroup$ – John Allen Mar 24 '17 at 16:11
  • $\begingroup$ The first term in the integral is $\int_0^1 2x^{1/2}\, dx = \left. \frac{2}{3/2} x^{3/2}\right|_{x=0}^{x=1} = \frac{4}{3}$. You can apply this formula to the other terms as well. $\endgroup$ – Connor Harris Mar 24 '17 at 16:14
  • $\begingroup$ Yes I see now. Thanks very much! $\endgroup$ – John Allen Mar 24 '17 at 16:15
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    $\begingroup$ @ConnorHarris But that would be integrating twice. The integration has already been done; the mistake was not to remove the integral sign. $\endgroup$ – Théophile Mar 24 '17 at 16:18
  • $\begingroup$ @Théophile Ooh, you're right. I guess John Allen's "sub in $x=1$" is correct (properly speaking, he wants the difference between $x=1$ and $x=0$, but the expression has value $0$ for $x=0$). $\endgroup$ – Connor Harris Mar 24 '17 at 16:22
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In derivative we subtract 1.

In integration we add 1.

Your solution is fine. Except last step.

$$=\int_0^1 (x^{-1/2} + 3x^{1/2} +5x^{5/2}) dx $$

$$=\left[2x^{1/2} + 2x^{3/2} + \frac{10}{7}x^{7/2}\right]_0^1$$

$$=\left[2\cdot1^{1/2} + 2\cdot1^{3/2} + \frac{10}{7}\cdot1^{7/2}\right]-\left[2\cdot0^{1/2} + 2\cdot0^{3/2} + \frac{10}{7}\cdot0^{7/2}\right]$$

Hope now you can proceed.

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    $\begingroup$ $$\int x^{\color{red}{n}}~dx=\frac{x^{n+1}}{n+1}+\color{red}{C}$$ $\endgroup$ – projectilemotion Mar 24 '17 at 16:25
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Putting $t=\sqrt{x}$ you have $dt=\frac{1}{2\sqrt{x}}dx$ and the limits stay the same. $$\int_0^12(1+3t^2+5t^6)$$

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Your work is correct, but for the last step write: $$ \left[ 2x^{1/2} + 2x^{3/2} + \frac{10}{7}x^{7/2}\right]_0^1 $$ because you have just done the integration as antiderivative: $F(x)=2x^{1/2} + 2x^{3/2} + \frac{10}{7}x^{7/2}$ and you have only to evaluate the primitive at the two limits of integration, so that your definite integrale is done by $F(1)-F(0)$.

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  • $\begingroup$ How can you say the work is correct when the final expression is blatantly wrong? I don't think it's good pedagogy to give the impression that it is just a matter of writing it differently. Precision is crucial in mathematics. $\endgroup$ – user21820 Mar 25 '17 at 9:02
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We put $t=\sqrt{x}$, so $dt=\frac{1}{2\sqrt{x}}dx$. $$\int_0^12(1+3t^2+5t^6)$$ $$=2\left[t+t^3+\frac{5t^7}{7}\right]^1_0$$ $$=2\left[\sqrt{x}+\sqrt{x}^3+\frac{5\sqrt{x}^7}{7}\right]^1_0$$ $$=\frac{38}{7}$$

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