Integration: Question 2

Worked solution

Part (a): Evaluate $\displaystyle\int_1^4 \left(3\sqrt{x} + \frac{4}{x^2}\right) dx$

Rewrite the integrand as powers of $x$:

$3\sqrt{x} + \frac{4}{x^2} = 3x^{1/2} + 4x^{-2}$

Integrate term by term:

$\int \left(3x^{1/2} + 4x^{-2}\right) dx = 3 \times \frac{x^{3/2}}{3/2} + 4 \times \frac{x^{-1}}{-1} = 2x^{3/2} - 4x^{-1} = 2x^{3/2} - \frac{4}{x}$

Evaluate at the limits:

At $x=4$: $\ 2(4)^{3/2} - \dfrac{4}{4} = 2(8) - 1 = 16 - 1 = 15$

At $x=1$: $\ 2(1)^{3/2} - \dfrac{4}{1} = 2 - 4 = -2$

$\int_1^4 \left(3\sqrt{x} + \frac{4}{x^2}\right) dx = 15 - (-2) = 17$

Recompute independently by differentiating the antiderivative: $\dfrac{d}{dx}\left(2x^{3/2}\right) = 3x^{1/2} = 3\sqrt{x}$, and $\dfrac{d}{dx}\left(-\dfrac{4}{x}\right) = \dfrac{4}{x^2}$. Both match the original integrand, so the antiderivative is correct and the value $17$ stands.

Part (b): Evaluate $\displaystyle\int_{-1}^{2} (3x-1)^3\,dx$

Using $\displaystyle\int (ax+b)^n\,dx = \frac{(ax+b)^{n+1}}{a(n+1)}+c$ with $a=3$, $n=3$:

$\int (3x-1)^3\,dx = \frac{(3x-1)^4}{3 \times 4} = \frac{(3x-1)^4}{12}$

Evaluate at the limits:

At $x=2$: $3(2)-1=5$, so $\dfrac{5^4}{12} = \dfrac{625}{12}$

At $x=-1$: $3(-1)-1=-4$, so $\dfrac{(-4)^4}{12} = \dfrac{256}{12}$

$\int_{-1}^{2} (3x-1)^3\,dx = \frac{625}{12} - \frac{256}{12} = \frac{369}{12} = \frac{123}{4}$

Recompute independently using the substitution $u = 3x-1$, so $du = 3\,dx$: when $x=2$, $u=5$; when $x=-1$, $u=-4$.

$\int u^3 \cdot \frac{du}{3} = \frac{u^4}{12}$

At $u=5$: $\dfrac{625}{12}$. At $u=-4$: $\dfrac{256}{12}$. Difference $= \dfrac{369}{12} = \dfrac{123}{4}$, exactly matching the first method, confirming the result.

Final answers

• (a) $\displaystyle\int_1^4 \left(3\sqrt{x} + \frac{4}{x^2}\right) dx = 17$
• (b) $\displaystyle\int_{-1}^{2} (3x-1)^3\,dx = \dfrac{123}{4}\ (=30.75)$