Question 1
The graphs of $f(x)=e^x$ and $g(x)=x^2-1$ are shown.
The area of the shaded section bounded by these graphs between the lines $x=0$ and $x=1$ is
Not quite. $1-e$ is negative, about $-1.718$, and an area can’t be negative. Integrate the top curve minus the bottom curve, from the lower bound to the upper bound, and the result comes out positive.
Not quite. $e-2$ is what you get if you write $2x$ as the antiderivative of $x^2$: $\bigl[e^x-2x+x\bigr]_0^1=e-2$. But $2x$ is the derivative of $x^2$. The same number also comes from $\int_0^1\left(e^x-1\right)dx$, which drops the $x^2$ and flips the sign of the 1. The antiderivative of $x^2$ is $\dfrac{x^3}{3}$.
Close. $e-\dfrac53$ is $\int_0^1\left(e^x+x^2-1\right)dx$, what you get by adding $g(x)$ instead of subtracting it. It’s the same as taking the $\dfrac23$ below the $x$-axis away from the $e-1$ above it. The part below the axis adds to the area: $e-1+\dfrac23=e-\dfrac13$.
That’s it. Step through the hints below to see why.
Nothing is shaded yet. The region starts at the line $x=0$, where the strip runs from $g(0)=-1$ up to $f(0)=1$.
The black curve stays above the blue one the whole way, so every strip has height $e^x-\left(x^2-1\right)$. Keep going to the line $x=1$.
This is the shaded section from the question. Adding up every strip from $x=0$ to $x=1$ gives $e-\dfrac13\approx2.385$.
The light part is above the $x$-axis, under the black curve. The darker part is below the axis, above the blue curve. Both parts are area, so both add to the total.
Above the axis is $e-1$, and below it is $\dfrac23$. Together they make $e-\dfrac13$. Taking the $\dfrac23$ away instead of adding it gives $e-\dfrac53$, option (C).
Decide which graph is on top between $x=0$ and $x=1$.
$f(0)=1,\ g(0)=-1$$f(1)=e,\ g(1)=0$
$f(x)=e^x$ is on top the whole way, so the area is the integral of top minus bottom.
$\displaystyle A=\int_0^1\Bigl(e^x-\left(x^2-1\right)\Bigr)dx$
Part of the region is below the $x$-axis. Top minus bottom handles that for you, so there’s no need to split it.
Expand the bracket, then integrate term by term.
$\displaystyle A=\int_0^1\left(e^x-x^2+1\right)dx$$\displaystyle =\Bigl[e^x-\frac{x^3}{3}+x\Bigr]_0^1$
Watch the sign: subtracting $x^2-1$ turns the $-1$ into $+1$.
Substitute the top bound, then subtract the bottom one. Don’t drop the lower bound: $e^0=1$, not 0.
$=\left(e-\dfrac13+1\right)-\left(1-0+0\right)$$=e-\dfrac13$
A quick check: $e-\dfrac13\approx2.385$. The strip is 2 tall at $x=0$ and $e\approx2.718$ tall at $x=1$, and the region is 1 wide, so an area a bit over 2 makes sense.
On $0\le x\le1$, $e^x$ is above $x^2-1$, so integrate top minus bottom.
$$\begin{aligned}A&=\int_0^1\left(e^x-x^2+1\right)dx\\&=\Bigl[e^x-\frac{x^3}{3}+x\Bigr]_0^1\\&=\left(e-\frac13+1\right)-1\\&=e-\frac13\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 Mathematical Methods 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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