Area under a curve
\[A=\int_a^b y\,dx\quad\text{or}\quad A=\int_c^d x\,dy\]
Applications of Integrals
If \(f(x)\ge0\) on \([a,b]\), the area of the region bounded by \(y=f(x)\), the \(x\)-axis and the lines \(x=a,\ x=b\) is
\[A=\int_a^bf(x)\,dx=\int_a^by\,dx.\]
Think of the region as made of thin vertical strips of height \(y\) and width \(dx\); integration adds them up. Similarly, with horizontal strips, the area between \(x=g(y)\), the \(y\)-axis and \(y=c,\ y=d\) is \(\int_c^dx\,dy\).
If \(f(x)\le0\), \(\int_a^bf\) is negative; the area is \(\left|\int_a^bf\right|\). If \(f\) changes sign, split the interval at its zeros and add the absolute values.
\[A=\int_a^b y\,dx\quad\text{or}\quad A=\int_c^d x\,dy\]
\[\int_0^a\sqrt{a^2-x^2}\,dx=\frac{\pi a^2}{4},\qquad \text{ellipse area}=\pi ab\]
\[y^2=4ax,\ x=a:\quad A=\frac{8a^2}{3}\]
A circle or ellipse centred at the origin is symmetric in both axes: compute the first-quadrant area and multiply by 4. A parabola y² = 4ax is symmetric about the x-axis: double the upper half.
\(\int_0^{2\pi}\sin x\,dx=0\), but the area between \(\sin x\) and the axis on \([0,2\pi]\) is \(2+2=4\).
When using horizontal strips, the limits must be y-values.
Area in the first quadrant enclosed by \(\frac{x^2}{25}+\frac{y^2}{4}=1\).
\(y=\frac25\sqrt{25-x^2}\), so the area is \(\frac25\int_0^5\sqrt{25-x^2}\,dx=\frac25\cdot\frac{25\pi}{4}=\frac{5\pi}{2}\).
A fresh set of questions from this topic. Change answers freely – solutions appear after you submit.
Start PracticeSketch, choose strips, set limits, integrate. Always give a positive answer in square units.
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