Question 5
Determine $\displaystyle\int4x\left(3x^2+5\right)^3dx$
That’s it. Step through the hints below to see why.
Not quite. With $u=3x^2+5$, $du=6x\,dx$ and $4x\,dx=\frac23\,du$, so the integral is $\int\frac23u^3\,du$. (B) raises the power to $4$ but forgets to divide by the new power: $\frac23\cdot\frac{u^4}{4}=\frac16u^4$.
Not quite. The power has gone down from $3$ to $2$, which is the rule for differentiating: $\frac23\cdot3u^2=2u^2$. Integrating raises the power, so $\int\frac23u^3\,du=\frac23\cdot\frac{u^4}{4}+c$. Differentiating (C) gives $24x\left(3x^2+5\right)$, which isn’t the integrand.
Not quite. This comes from differentiating instead of integrating. The chain rule on the bracket, with $4x$ kept in front, gives $4x\cdot3\left(3x^2+5\right)^2\cdot6x$, which simplifies to $72x^2\left(3x^2+5\right)^2$. The question asks for the integral.
An antiderivative differentiates back to the integrand, so the right option’s derivative sits on the black curve at every $x$. A blue triangle at the top means the derivative is off the scale.
Differentiating (A) gives back $4x\left(3x^2+5\right)^3$ exactly, so the two curves coincide at every $x$. Option (A) is an antiderivative.
The derivative of (B) is $16x\left(3x^2+5\right)^3$, four times the integrand at every $x$. Option (B) raised the power to $4$ but never divided by $4$.
The derivative of (C) is $24x\left(3x^2+5\right)$, which stays below the integrand at every $x$ on the slider. In (C) the power went down from $3$ to $2$, which is what differentiating does.
The derivative of (D) is far above the integrand at every $x$ on the slider. Option (D) is what you get by differentiating the bracket and keeping $4x$ in front.
The bracket’s derivative is $6x$, and there is a multiple of $x$ outside it. That is the shape for a substitution.
$$u=3x^2+5,\qquad\frac{du}{dx}=6x$$
Replace $4x\,dx$. Since $du=6x\,dx$, $4x\,dx=\frac46\,du=\frac23\,du$.
$$\int4x\left(3x^2+5\right)^3dx=\int\frac23u^3\,du$$
Integrate: raise the power by 1 and divide by the new power. Then put $3x^2+5$ back.
$$\frac23\cdot\frac{u^4}{4}+c=\frac16\left(3x^2+5\right)^4+c$$
Check by differentiating: $\frac16\cdot4\left(3x^2+5\right)^3\cdot6x=4x\left(3x^2+5\right)^3$.
Let $u=3x^2+5$, so $du=6x\,dx$ and $4x\,dx=\frac23\,du$.
$$\begin{aligned}\int4x\left(3x^2+5\right)^3dx&=\int\frac23u^3\,du\\&=\frac23\cdot\frac{u^4}{4}+c\\&=\frac16\left(3x^2+5\right)^4+c\end{aligned}$$
Only mark this done when you could do it without help. Reading the solution does not count.
Question wording is from the 2020 QCAA Specialist Mathematics external assessment, © State of Queensland (QCAA) 2020, licensed under CC BY 4.0, and has been adapted. Tangent Tuition is not affiliated with the QCAA.
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