IB Maths AI HLFurther IntegrationPaper 1 & 2~6 min read
Area Between a Curve and the y-axis
To find the area trapped between a curve and the y-axis, you integrate with respect to y instead of x. Rearrange the equation to x = g(y), use y-limits, and evaluate ∫x dy. It’s the same idea as before, rotated a quarter-turn.
📘 What you need to know
Integrate in y: area between a curve and the y-axis is ∫cdx dy.
Rearrange first: make x the subject, x = g(y), before integrating.
Limits are y-values: c (lower) and d (upper) come from the y-axis, not the x-axis.
dy, not dx: every term integrates as a power of y.
Same sign rule: area left of the y-axis (x < 0) gives a negative integral — take the modulus.
Swap the variable, keep the method: find limits, integrate, substitute top minus bottom.
Turning it on its side
Area between a curve and the y-axisA = ∫cdx dy, where x = g(y)
✓ same definite-integral idea, with x and y swapped
Width measured horizontally
Each thin strip is horizontal: width x, height dy. Summing from y = c to y = d gives the area.
🤔 Why integrate with respect to y?
For the x-axis we add up vertical strips of height y and width dx. Against the y-axis the natural strips are horizontal: width x, thickness dy. Adding those strips up the y-direction means the variable of integration is y — so we need x written as a function of y.
🧠 “Make x the subject, integrate dy“
Whenever the boundary is the y-axis, rearrange to x = g(y) and use y-limits. If you find yourself with dx, you’ve reverted to the x-axis method by mistake.
Carrying it out
🧭 Recipe — area against the y-axis
Rearrange the curve to x = g(y).
Find the y-limitsc and d (from given values or where the curve meets the y-axis).
WriteA = ∫cdg(y) dy.
Integrate in y and substitute top minus bottom.
Modulus if the region lies left of the axis (negative x).
Finding the y-limits: a horizontal line y = 4 gives a limit directly; the x-axis gives y = 0; and where the curve meets the y-axis, set x = 0 and read off y.
Worked examples
WE 1
Find the area between y = x2, the y-axis, and y = 0 to y = 4 (first quadrant)
Rearrange to x = √y, then integrate in y.
x = √y = y1/2∫04 y1/2 dy = [23 y3/2]04= 23(8) − 0area = 163 square units
WE 2
Find the area between y = x3, the y-axis, from y = 0 to y = 8
Make x the subject: x = y1/3.
x = y1/3∫08 y1/3 dy = [34 y4/3]08= 34(16) − 0area = 12 square units
WE 3
Find the area between y = 2x + 1, the y-axis, from y = 1 to y = 5
Rearrange the line: x = y − 12.
x = y − 12∫15y − 12 dy = 12[y²2 − y]15= 12[(12.5 − 5) − (0.5 − 1)] = 12(8)area = 4 square units
WE 4
Find the area between y = ex, the y-axis, from y = 1 to y = e
Invert the exponential: x = ln y. (Recall ∫ ln y dy = y ln y − y.)
x = ln y∫1e ln y dy = [y ln y − y]1e= (e − e) − (0 − 1)area = 1 square unit
WE 5
Find the area between y = x2, the y-axis, from y = 1 to y = 4
Same curve as WE 1, but the strip now starts at y = 1.
x = √y = y1/2∫14 y1/2 dy = [23 y3/2]14= 23(8) − 23(1) = 23(7)area = 143 square units
💡 Top tips
Rearrange to x = g(y) before doing anything else.
Limits are y-values — read them off the y-axis.
Integrate in y: every term a power of y, ending in dy.
Sketch sideways — picture horizontal strips of width x.
Modulus if the region sits to the left of the y-axis.
GDC still evaluates it — just feed it the function of y.
⚠ Common mistakes
Integrating in x out of habit when the boundary is the y-axis.
Forgetting to rearrange to x = g(y) first.
Using x-limits instead of y-limits.
Slips inverting the function (e.g. √ vs cube root, ln vs e).
Ignoring the sign when the curve is left of the axis.
Next up — Areas Between Curves. You’ve now found areas against both axes — vertical strips (dx) for the x-axis, horizontal strips (dy) for the y-axis. The next topic puts two curves together: the area enclosed between them is the integral of (top − bottom) across their points of intersection, and the same strip-thinking tells you which way round to subtract.
Need help with Further Integration?
Get 1-on-1 help from an IB examiner who knows exactly what Paper 1 & 2 are looking for.
