Integration — practice questions

The area is given by the definite integral $A = \int_0^2 x\sqrt{x^2+5} \, dx$.

This is not a standard integral, so we use substitution. The expression inside the square root is a good candidate for substitution.

Let $u = x^2+5$. Then, differentiating with respect to $x$: $\frac{du}{dx} = 2x$. Rearranging gives $du = 2x \, dx$, or $x \, dx = \frac{1}{2} du$.

Now, we must change the limits of integration from $x$-values to $u$-values: When $x=0$, $u = 0^2+5 = 5$. When $x=2$, $u = 2^2+5 = 9$.

Substitute everything into the integral: $A = \int_5^9 \sqrt{u} \, (\frac{1}{2} du) = \frac{1}{2} \int_5^9 u^{1/2} \, du$.

Now integrate with respect to $u$: $A = \frac{1}{2} \left[ \frac{u^{3/2}}{3/2} \right]_5^9 = \frac{1}{2} \left[ \frac{2}{3}u^{3/2} \right]_5^9 = \frac{1}{3} [u^{3/2}]_5^9$.

Finally, substitute the new limits: $A = \frac{1}{3} (9^{3/2} - 5^{3/2}) = \frac{1}{3} ((\sqrt{9})^3 - 5\sqrt{5}) = \frac{1}{3} (3^3 - 5\sqrt{5})$.

The exact area is $\frac{1}{3}(27 - 5\sqrt{5})$.

[M1] for choosing a suitable substitution and finding $\frac{du}{dx}$. [A1] for correct substitution including converting $x \, dx$. [M1] for changing the limits correctly. [A1] for the correct integral in terms of $u$. [M1] for integrating $u^{1/2}$. [A1] for the final exact answer.

We have a product of an algebraic function ($x^2$) and a logarithmic function ($\ln x$). We use integration by parts.

Using the LIATE principle, we choose the Logarithmic function as $u$. Let $u = \ln x$ and $\frac{dv}{dx} = x^2$.

Then we differentiate $u$ and integrate $\frac{dv}{dx}$: $\frac{du}{dx} = \frac{1}{x}$ and $v = \int x^2 dx = \frac{x^3}{3}$.

Apply the integration by parts formula: $\int u \frac{dv}{dx} dx = uv - \int v \frac{du}{dx} dx$. $\int_1^e x^2 \ln x \, dx = \left[ (\ln x) \left(\frac{x^3}{3}\right) \right]_1^e - \int_1^e \left(\frac{x^3}{3}\right) \left(\frac{1}{x}\right) dx$.

Simplify the expression: $= \left[ \frac{x^3 \ln x}{3} \right]_1^e - \int_1^e \frac{x^2}{3} dx$.

Now, evaluate the remaining integral: $= \left[ \frac{x^3 \ln x}{3} \right]_1^e - \left[ \frac{x^3}{9} \right]_1^e$.

Substitute the limits. Remember $\ln e = 1$ and $\ln 1 = 0$. $= \left( \frac{e^3 \ln e}{3} - \frac{1^3 \ln 1}{3} \right) - \left( \frac{e^3}{9} - \frac{1^3}{9} \right)$. $= \left( \frac{e^3(1)}{3} - \frac{1(0)}{3} \right) - \left( \frac{e^3}{9} - \frac{1}{9} \right)$. $= \frac{e^3}{3} - 0 - \frac{e^3}{9} + \frac{1}{9}$.

Combine the terms: $= \frac{3e^3}{9} - \frac{e^3}{9} + \frac{1}{9} = \frac{2e^3 + 1}{9}$.

[M1] for correct choice of $u$ and $\frac{dv}{dx}$. [A1] for correct $\frac{du}{dx}$ and $v$. [M1] for correct application of the by parts formula. [A1] for correct integration of the second term. [M1] for substituting limits correctly. [A1] for the final exact answer.