Application of Integrals: questions with answers

The part of the ellipse between $x = \pm\dfrac{a}{2}$ is $\dfrac{4b}{a}\int_0^{a/2}\sqrt{a^2 - x^2}\,dx$, with both terms of the anti derivative alive at $x = \dfrac{a}{2}$ ($\sin^{-1}\dfrac{1}{2} = \dfrac{\pi}{6}$): $9\sqrt{3} + 6\pi$. The strip is not half of $\pi ab$ -- an ellipse is not a rectangle -- and without the factor $\dfrac{b}{a}$ the integral belongs to the circle of radius $a$.

Line 1 is correct. Line 2 is the first wrong line: the factor $\dfrac{ 2 }{ 4 }$ has been dropped, so the integral measures a strip of the circle of radius $4$. Line 3 evaluates that integral correctly. The area is $8\left(\dfrac{\pi}{3} + \dfrac{\sqrt{3}}{2}\right)$.

Line 1 is correct. Line 2 is the first wrong line: the $x$-axis is one of the boundaries, so the region lies on one side of it and is bounded by ONE branch, $y = 2\sqrt{x}$. Line 3 evaluates the doubled integral correctly. The area is $4/3$.

Line 1 is the first wrong line: the area under a curve is the integral of $y$, a sum of strips $y\,dx$, not the difference of two heights. Lines 2 and 3 compute that difference correctly. $A = \int_{ 2 }^{ 3 } x^2\,dx = 19/3$.

Line 1 is correct. Line 2 is the first wrong line: $4y$ is $x^2$, not $x$; the strip length needs the curve solved for $x$, which is $2\sqrt{y}$. Line 3 evaluates the wrong integral correctly. The area is $8/3$.

Line 1 is correct. Line 2 is the first wrong line: the integrand is the strip length $x$, which the curve gives as $x = \dfrac{y^2}{ 4 }$; $y$ is the height of the strip above the axis, not its length. Line 3 evaluates the wrong integral correctly. The area is $18$.

Line 1 is the first wrong line: the limits are where the curves MEET, and solving $\dfrac{x^2}{ 4 } = 4x$ gives $x = 0$ and $x = 16$, not $4$. Lines 2 and 3 integrate the right expression over the wrong interval. The area is $512/3$.

Lines 1 and 2 are correct. Line 3 is the first wrong line: the left part must equal HALF the whole area, and it has been set equal to the whole -- which is why the answer lands on the far edge of the region. With half the total, $a = \left(\dfrac{27}{2}\right)^{2/3}$.

Each piece of the region has a positive area: the piece below the axis has area $|A_1|$ and the piece above has area $A_2$, and the areas of the two pieces ADD. $A_2 - |A_1|$ is the single integral $\int_p^q f(x)\,dx$ in disguise, in which the piece below the axis cancels part of the piece above.

The curve crosses the $x$-axis at $x = 0$, inside the interval, and lies below the axis to the left of it. The piece below has a negative integral, so the area is the absolute value of that piece plus the piece above: $\dfrac{16}{3}$. The single integral straight across lets the two pieces cancel.

The line $y = x$ through the centre cuts the quadrant in two; the region below it is a triangle under the line up to the intersection plus the piece under the arc beyond it: $\int_0^{a/\sqrt{2}} x\,dx + \int_{a/\sqrt{2}}^{a}\sqrt{a^2 - x^2}\,dx = \dfrac{9\pi}{2}$, one eighth of the circle. Integrating the arc from $0$ to $a$ ignores the line and gives the whole quadrant.

The segment lies between the cutting line and the circle, so the integral runs from the line to $x = a$ and the anti derivative is evaluated at BOTH limits; the factor $2$ counts the halves above and below the $x$-axis: $\dfrac{25\pi}{3} - \dfrac{25\sqrt{3}}{4}$. Evaluating at the upper limit alone gives the whole half circle, and the segment is not its share of the diameter times $\pi a^2$.

The graph may have an axis or a corner, but the INTERVAL is not symmetric about it, so the two pieces are different shapes and both must be computed: $\dfrac{25}{2}$. Doubling one piece measures a region the question did not ask about.

The curve is odd, so its INTEGRAL over the symmetric interval is $0$ -- the piece below the axis cancels the piece above. But the two pieces have equal, positive AREA, so the area is twice one piece: $\dfrac{1}{2}$. One piece alone is only half of the region.

Line 1 is correct. Line 2 is the first wrong line: the graph is $|x + 1|$ lifted by $3$, so each arm is $3 - (x + 1)$ and $3 + (x + 1)$; the constant has been dropped and the rectangle $3 \times 7$ under the V is lost. Line 3 evaluates the wrong integrals correctly. The area is $67/2$.