Evaluate the integral: $\int_0^1 \dfrac{r^3}{\sqrt{16 + r^2}} \, dr$ 

Neetesh Kumar | September 20, 2024

This is the solution to Integral by Parts homework question
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Step-by-Step Solution:

Evaluate the integral: $\int_0^1 \frac{r^3}{\sqrt{16 + r^2}} \, dr$

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Substitute:

Let $u = 16 + r^2$, simplifying the denominator's square root.

Then, the differential $du$ is: $du = 2r \, dr$ So: $r \, dr = \frac{du}{2}$

Rewriting the integral:

The limits of integration change when substituting $u$:

• When $r = 0$, $u = 16$,
• When $r = 1$, $u = 17$.

Now, rewrite the integral in terms of $u$: $\int_{16}^{17} \frac{(u - 16)}{\sqrt{u}} \cdot \frac{du}{2}$

Simplifying the expression:

Factor out constants: $\frac{1}{2} \int_{16}^{17} \frac{u - 16}{\sqrt{u}} \, du$

Split the integral:

$\frac{1}{2} \left( \int_{16}^{17} \frac{u}{\sqrt{u}} \, du - 16 \int_{16}^{17} \frac{1}{\sqrt{u}} \, du \right)$ Simplify the powers of $u$: $\frac{1}{2} \left( \int_{16}^{17} u^{1/2} \, du - 16 \int_{16}^{17} u^{-1/2} \, du \right)$

Evaluate each integral:

• The first integral: $\int u^{1/2} \, du = \frac{2}{3} u^{3/2}$

• The second integral: $\int u^{-1/2} \, du = 2u^{1/2}$

Substitute back the limits:

• For the first term: $\frac{2}{3} \left( 17^{3/2} - 16^{3/2} \right)$

• For the second term: $16 \cdot 2 \left( 17^{1/2} - 16^{1/2} \right)$

After putting the values back

• $\frac{1}{3} \left( 17^{3/2} - 64 \right) - 16 \left( 17^{1/2} - 4 \right)$

After substituting the values and simplifying the expression, the final result for the integral is:

• the value of integral = $\frac{128-31\sqrt{17}}{3} = 0.061242$

Final answer:

$\boxed{\frac{128-31\sqrt{17}}{3} \ or \ 0.061242}$

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