Anti-derivatives

Problem. Evaluate \int \left(3x^{2} - \frac{4}{x} + 5\sqrt{x} + e^{x}\right)dx.

Solution. Split the integral term by term and apply the table to each.

For 5\sqrt{x} = 5x^{1/2} the power rule gives exponent \frac12 + 1 = \frac32:

and \int e^{x}dx = e^{x}. Collecting, and writing one constant for all four,

Check. Differentiate: 3x^{2} - \frac4x + \frac{10}{3}\cdot\frac32 x^{1/2} + e^{x} = 3x^{2} - \frac4x + 5\sqrt{x} + e^{x}. Correct.

Problem. Evaluate \int \left(\sqrt{x} + \frac{1}{\sqrt{x}}\right)^{2}dx and \int \frac{(x^{2}+1)^{2}}{x^{2}}\,dx.

Solution. Neither integrand is in the table, but both become table entries after ordinary algebra. Always expand or divide out first.

For the second, expand the numerator and divide each term by x^{2}:

Problem. Evaluate (a) \int \tan^{2}x\,dx, (b) \int \sin^{2}x\,dx, (c) \int \frac{dx}{1+\cos x}.

Solution. Every one of these is solved by a trigonometric identity, not by a new integration rule.

(a) Use \sec^{2}x - \tan^{2}x = 1, so \tan^{2}x = \sec^{2}x - 1:

(b) Use the power-reduction identity \cos 2x = 1 - 2\sin^{2}x, so \sin^{2}x = \frac{1 - \cos 2x}{2}:

(c) Use 1 + \cos x = 2\cos^{2}\frac{x}{2}:

Problem. Evaluate (a) \int (2x+3)^{5}dx, (b) \int \frac{dx}{3x+2}, (c) \int \sin(5x-2)\,dx, (d) \int \frac{dx}{\sqrt{2x+1}}.

Solution. All four are of the form f(ax+b), so integrate as if the bracket were x and divide by a.

(a) a = 2: \quad \int (2x+3)^{5}dx = \frac{1}{2}\cdot\frac{(2x+3)^{6}}{6} + C = \frac{(2x+3)^{6}}{12} + C

(b) a = 3: \quad \int \frac{dx}{3x+2} = \frac{1}{3}\ln|3x+2| + C

(c) a = 5: \quad \int \sin(5x-2)\,dx = -\frac{1}{5}\cos(5x-2) + C

(d) a = 2 and \int x^{-1/2}dx = 2\sqrt{x}: \quad \int \frac{dx}{\sqrt{2x+1}} = \frac{1}{2}\cdot 2\sqrt{2x+1} + C = \sqrt{2x+1} + C

Problem. Evaluate \int x\sqrt{x^{2}+1}\,dx.

Solution. The derivative of x^{2}+1 is 2x, and an x is already present.

Put u = x^{2}+1, so du = 2x\,dx, i.e. x\,dx = \frac{du}{2}.

Put u back: \quad = \frac{1}{3}\left(x^{2}+1\right)^{3/2} + C.

Check. \frac{d}{dx}\frac{(x^{2}+1)^{3/2}}{3} = \frac{1}{3}\cdot\frac32(x^{2}+1)^{1/2}\cdot 2x = x\sqrt{x^{2}+1}. Correct.

Problem. Prove that \int \tan x\,dx = \ln|\sec x| + C and \int \cot x\,dx = \ln|\sin x| + C.

Solution. Write \tan x = \dfrac{\sin x}{\cos x} and put u = \cos x, so du = -\sin x\,dx:

For the cotangent, u = \sin x gives du = \cos x\,dx and the numerator is already \cos x:

Problem. Evaluate (a) \int \frac{e^{\tan^{-1}x}}{1+x^{2}}\,dx, (b) \int \frac{x^{3}}{x^{4}+5}\,dx, (c) \int \frac{dx}{x\ln x}, (d) \int \sin^{3}x\cos x\,dx.

Solution. In each case spot the inside function and its derivative.

(a) u = \tan^{-1}x, du = \frac{dx}{1+x^{2}}: \quad \int e^{u}du = e^{u} + C = e^{\tan^{-1}x} + C

(b) u = x^{4}+5, du = 4x^{3}dx: \quad \frac14\int\frac{du}{u} = \frac14\ln\left|x^{4}+5\right| + C

(c) u = \ln x, du = \frac{dx}{x}: \quad \int\frac{du}{u} = \ln\left|\ln x\right| + C

(d) u = \sin x, du = \cos x\,dx: \quad \int u^{3}du = \frac{u^{4}}{4} + C = \frac{\sin^{4}x}{4} + C

Problem. Prove the standard result \int \sec x\,dx = \ln|\sec x + \tan x| + C.

Solution. The trick is to multiply and divide by \sec x + \tan x, which manufactures a "derivative over function" pattern.

Now put u = \sec x + \tan x. Then du = \left(\sec x\tan x + \sec^{2}x\right)dx, which is exactly the numerator:

The same trick with \csc x - \cot x gives \int \csc x\,dx = \ln|\csc x - \cot x| + C.

Problem. Evaluate \int x e^{x}\,dx.

Solution. Algebraic before Exponential, so u = x and dv = e^{x}dx.

Then du = dx and v = e^{x}.

Check. \frac{d}{dx}\left[(x-1)e^{x}\right] = e^{x} + (x-1)e^{x} = x e^{x}. Correct.

Had we chosen u = e^{x} instead, we would have got \frac{x^{2}}{2}e^{x} - \int \frac{x^{2}}{2}e^{x}dx — a harder integral. That is what ILATE protects you from.

Problem. Evaluate \int x^{2}\sin x\,dx (the repeated case).

Solution. Algebraic before Trigonometric: u = x^{2}, dv = \sin x\,dx, so du = 2x\,dx and v = -\cos x.

The new integral is still a product, but the power of x has dropped by one. Apply parts again with u = x, dv = \cos x\,dx, v = \sin x:

Substituting back,

Check. Differentiating gives -2x\cos x + x^{2}\sin x + 2\sin x + 2x\cos x - 2\sin x = x^{2}\sin x. Correct. An integrand x^{n}\sin x needs exactly n applications of parts.

Problem. Evaluate \int \ln x\,dx, \int x^{2}\ln x\,dx and \int \tan^{-1}x\,dx.

Solution. A lone logarithm or inverse-trig function is integrated against the invisible factor 1.

(a) u = \ln x, dv = 1\,dx, so du = \frac{dx}{x} and v = x:

(b) L before A: u = \ln x, dv = x^{2}dx, so v = \frac{x^{3}}{3}:

(c) u = \tan^{-1}x, dv = 1\,dx, so du = \frac{dx}{1+x^{2}} and v = x:

using \int\frac{g'}{g} = \ln|g| with g = 1+x^{2} (always positive, so no modulus is needed).

Problem. Evaluate I = \int e^{x}\sin x\,dx (the cyclic case).

Solution. Neither factor simplifies on differentiation, so parts will never terminate. Instead, apply it twice and recover the original integral.

Take u = \sin x, dv = e^{x}dx, so v = e^{x}:

Apply parts to the new integral with u = \cos x, dv = e^{x}dx:

Substituting back,

The general result, obtained the same way, is

For example \int e^{2x}\cos 3x\,dx = \frac{e^{2x}}{13}\left(2\cos 3x + 3\sin 3x\right) + C.

Problem. Evaluate (a) \int \frac{dx}{4x^{2}+9}, (b) \int \frac{dx}{x^{2}-16}, (c) \int \frac{dx}{\sqrt{9-4x^{2}}}.

Solution. Force each into the shape x^{2} \pm a^{2} by taking the leading coefficient outside.

(a) 4x^{2}+9 = 4\left(x^{2} + \frac94\right), so a = \frac32:

(b) Here a = 4 directly:

(c) 9-4x^{2} = 4\left(\frac94 - x^{2}\right), so \sqrt{9-4x^{2}} = 2\sqrt{\left(\frac32\right)^{2}-x^{2}}:

Problem. Evaluate (a) \int \frac{dx}{x^{2}+6x+13}, (b) \int \frac{dx}{\sqrt{x^{2}+4x+13}}, (c) \int \frac{dx}{\sqrt{5-4x-x^{2}}}.

Solution. Complete the square in every case.

(a) x^{2}+6x+13 = (x+3)^{2} + 4 = (x+3)^{2} + 2^{2}. Put t = x+3, dt = dx:

(b) x^{2}+4x+13 = (x+2)^{2} + 9 = (x+2)^{2} + 3^{2}. With t = x+2:

(c) Take out the minus sign first: 5-4x-x^{2} = 5 - \left(x^{2}+4x\right) = 5 + 4 - (x+2)^{2} = 3^{2} - (x+2)^{2}. With t = x+2:

Problem. Evaluate \int \frac{2x+3}{x^{2}+4x+5}\,dx.

Solution. The derivative of x^{2}+4x+5 is 2x+4. Write

so the integral splits into two:

The first is \frac{g'}{g}, giving \ln\left|x^{2}+4x+5\right|; and since x^{2}+4x+5 = (x+2)^{2}+1 > 0 we may drop the modulus. For the second, t = x+2 gives \int \frac{dt}{t^{2}+1} = \tan^{-1}(x+2). Hence

Problem. Evaluate \int \frac{x+2}{\sqrt{x^{2}+2x+3}}\,dx.

Solution. The derivative of x^{2}+2x+3 is 2x+2, so

In the first integral put u = x^{2}+2x+3, du = (2x+2)dx:

For the second, x^{2}+2x+3 = (x+1)^{2}+2, so with t = x+1, \int\frac{dt}{\sqrt{t^{2}+2}} = \ln\left|t+\sqrt{t^{2}+2}\right|. Altogether

Problem. Derive \int \sqrt{a^{2}-x^{2}}\,dx = \frac{x}{2}\sqrt{a^{2}-x^{2}} + \frac{a^{2}}{2}\sin^{-1}\frac{x}{a} + C.

Solution. Put x = a\sin\theta, so dx = a\cos\theta\,d\theta and \sqrt{a^{2}-x^{2}} = a\cos\theta.

Use \cos^{2}\theta = \frac{1+\cos 2\theta}{2}:

From the first reference triangle, \theta = \sin^{-1}\frac{x}{a}, \sin\theta = \frac{x}{a} and \cos\theta = \frac{\sqrt{a^{2}-x^{2}}}{a}, so

Problem. Evaluate \int \frac{dx}{x^{2}\sqrt{x^{2}-9}}.

Solution. The radical is \sqrt{x^{2}-a^{2}} with a = 3, so put x = 3\sec\theta, dx = 3\sec\theta\tan\theta\,d\theta and \sqrt{x^{2}-9} = 3\tan\theta.

From the third reference triangle, the hypotenuse is x, the adjacent side is 3, and the opposite side is \sqrt{x^{2}-9}, so \sin\theta = \frac{\sqrt{x^{2}-9}}{x}. Hence

Check. Differentiating \frac{1}{9}\left(1 - \frac{9}{x^{2}}\right)^{1/2}, written this way to make the algebra short, returns \frac{1}{x^{2}\sqrt{x^{2}-9}}. Correct.

Problem. Evaluate \int \frac{dx}{x\sqrt{1+x^{2}}} by the inverse substitution x = \frac{1}{t}.

Solution. When a single power of x sits outside a heavy radical, the substitution x = \frac1t turns the radical inside out. Here dx = -\frac{dt}{t^{2}} and

Taking t>0 (i.e. x>0),

Putting t = \frac1x back,

Problem. Evaluate \int \frac{dx}{5+4\cos x}.

Solution. Put t = \tan\frac{x}{2}. Then

and dx = \frac{2\,dt}{1+t^{2}}, so the factor 1+t^{2} cancels completely:

Problem. Evaluate \int \frac{dx}{3\sin x + 4\cos x}.

Solution. Again put t = \tan\frac{x}{2}:

Split by partial fractions. Writing \frac{1}{(2t+1)(t-2)} = \frac{A}{2t+1} + \frac{B}{t-2} gives 1 = A(t-2) + B(2t+1). Put t=2: 1 = 5B, so B = \frac15. Put t = -\frac12: 1 = -\frac52 A, so A = -\frac25. Hence

Notice the coefficient \frac15 = \frac{1}{\sqrt{3^{2}+4^{2}}}. That is no accident: a\sin x + b\cos x = \sqrt{a^{2}+b^{2}}\,\sin(x+\alpha), so the integral always carries a factor \frac{1}{\sqrt{a^{2}+b^{2}}} — a fast way to check your answer.

Problem. Evaluate \int \frac{dx}{(x-1)(x+2)} — distinct linear factors.

Solution. Write

Multiply through by (x-1)(x+2): \quad 1 = A(x+2) + B(x-1).

Put x = 1: \;1 = 3A, so A = \frac13. Put x = -2: \;1 = -3B, so B = -\frac13. Therefore

Problem. Evaluate \int \frac{dx}{(x-1)(x-2)(x-3)} and \int \frac{x\,dx}{(x-1)(x-2)}.

Solution. (a) With \frac{1}{(x-1)(x-2)(x-3)} = \frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x-3} we get 1 = A(x-2)(x-3) + B(x-1)(x-3) + C(x-1)(x-2).

Put x=1: 1 = A(-1)(-2) = 2A, so A = \frac12. Put x=2: 1 = B(1)(-1) = -B, so B = -1. Put x=3: 1 = C(2)(1) = 2C, so C = \frac12.

(b) \frac{x}{(x-1)(x-2)} = \frac{A}{x-1} + \frac{B}{x-2} gives x = A(x-2)+B(x-1). Put x=1: A = -1. Put x=2: B = 2. So

Problem. Evaluate \int \frac{3x+5}{(x-1)^{2}(x+1)}\,dx — a repeated linear factor.

Solution. The repeated factor needs two terms:

Multiplying by (x-1)^{2}(x+1):

Put x=1: \;8 = 2B, so B = 4. Put x=-1: \;2 = 4D, so D = \frac12. Compare the coefficients of x^{2}: \;0 = A + D, so A = -\frac12.

using \int (x-1)^{-2}dx = -\frac{1}{x-1}. Note that only the first-power terms give logarithms.

Problem. Evaluate \int \frac{dx}{(x-1)\left(x^{2}+1\right)} — an irreducible quadratic factor.

Solution. Since x^{2}+1 has no real factors, it takes a linear numerator:

Put x=1: \;1 = 2A, so A = \frac12. Coefficient of x^{2}: \;0 = A + B, so B = -\frac12. Constant term: \;1 = A - D, so D = -\frac12.

Split the second integral: \frac{x}{x^{2}+1} is a \frac{g'}{2g} pattern and \frac{1}{x^{2}+1} is a standard form.

Problem. Evaluate \int \frac{x^{3}+2x+1}{\left(x^{2}+1\right)^{2}}\,dx — a repeated quadratic factor.

Solution. The decomposition is

Multiplying up, \;x^{3}+2x+1 = (Ax+B)\left(x^{2}+1\right) + Dx + E, i.e.

Comparing coefficients: A = 1, B = 0, A + D = 2 so D = 1, and B + E = 1 so E = 1. Hence

The first two use u = x^{2}+1, du = 2x\,dx:

The third needs x = \tan\theta, dx = \sec^{2}\theta\,d\theta:

With \theta = \tan^{-1}x, \sin\theta = \frac{x}{\sqrt{1+x^{2}}} and \cos\theta = \frac{1}{\sqrt{1+x^{2}}}, this is \frac12\tan^{-1}x + \frac{x}{2\left(1+x^{2}\right)}. Collecting everything,

Problem. Evaluate \int \frac{x^{2}+1}{(x+1)(x+2)}\,dx and \int \frac{x^{2}}{x^{2}+1}\,dx — improper fractions.

Solution. In both, the numerator degree is not less than the denominator degree, so divide first.

(a) \frac{x^{2}+1}{x^{2}+3x+2} = 1 + \frac{-3x-1}{(x+1)(x+2)}. Now \frac{-3x-1}{(x+1)(x+2)} = \frac{A}{x+1} + \frac{B}{x+2} gives -3x-1 = A(x+2)+B(x+1). Put x=-1: 2 = A. Put x=-2: 5 = -B, so B=-5.

(b) \frac{x^{2}}{x^{2}+1} = 1 - \frac{1}{x^{2}+1}, so the integral is x - \tan^{-1}x + C at once. Dividing out first turns a hard-looking problem into a one-line one.

Problem. Evaluate \int \frac{dx}{x^{3}+1}.

Solution. Factorise the denominator using a^{3}+b^{3} = (a+b)\left(a^{2}-ab+b^{2}\right): \;x^{3}+1 = (x+1)\left(x^{2}-x+1\right), and x^{2}-x+1 is irreducible (discriminant 1-4 = -3 < 0). So

Put x=-1: \;1 = 3A, so A = \frac13. Coefficient of x^{2}: 0 = A + B, so B = -\frac13. Constant: 1 = A + D, so D = \frac23. Then

For the second integral, split the numerator using (x^{2}-x+1)' = 2x-1: \;x-2 = \frac12(2x-1) - \frac32. Also x^{2}-x+1 = \left(x-\frac12\right)^{2} + \frac34, so

Therefore

Problem. Evaluate (a) \int_{1}^{2}\left(x^{2}+1\right)dx, (b) \int_{0}^{\pi}\sin x\,dx, (c) \int_{0}^{3\pi/2}\sin x\,dx, (d) \int_{0}^{2\pi}\sin x\,dx.

Solution. Find an anti-derivative, then subtract.

(a) \left[\frac{x^{3}}{3}+x\right]_{1}^{2} = \left(\frac83+2\right) - \left(\frac13+1\right) = \frac{14}{3} - \frac43 = \frac{10}{3}

(b) \left[-\cos x\right]_{0}^{\pi} = -\cos\pi + \cos 0 = 1+1 = 2

(c) \left[-\cos x\right]_{0}^{3\pi/2} = -\cos\frac{3\pi}{2} + \cos 0 = 0 + 1 = 1

(d) \left[-\cos x\right]_{0}^{2\pi} = -1+1 = 0

Parts (b), (c) and (d) are the three cases drawn in the figure: all area above, part below, and exact cancellation.

Problem. Evaluate \int_{0}^{4}\frac{x}{\sqrt{x^{2}+9}}\,dx and \int_{0}^{1} x e^{x}\,dx.

Solution. (a) Put u = x^{2}+9, so du = 2x\,dx. When x=0, u=9; when x=4, u=25.

(b) By parts, using Worked example 13.9:

Problem. Evaluate I = \displaystyle\int_{0}^{\pi/2} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\,dx.

Solution. No anti-derivative of this integrand exists in elementary terms, so use P4 with a = \frac{\pi}{2}. Since \sin\left(\frac{\pi}{2}-x\right) = \cos x and \cos\left(\frac{\pi}{2}-x\right) = \sin x,

Add the two expressions for I. The denominators are now identical:

The same argument gives \int_{0}^{\pi/2}\frac{dx}{1+\tan x} = \frac{\pi}{4} and \int_{0}^{\pi/2}\sin^{2}x\,dx = \int_{0}^{\pi/2}\cos^{2}x\,dx = \frac{\pi}{4}, because in each case the two copies add to 1.

Problem. Use P6 to evaluate \int_{-1}^{1}\left(x^{3}+x\right)dx and \int_{-2}^{2}x^{2}\,dx.

Solution. Test the parity first.

(a) f(-x) = -x^{3}-x = -f(x), so f is odd and the integral over a symmetric interval is 0. (Check directly: \left[\frac{x^{4}}{4}+\frac{x^{2}}{2}\right]_{-1}^{1} = \frac34 - \frac34 = 0.)

(b) f(-x) = x^{2} = f(x), so f is even:

For a mixed integrand, split it: \int_{-2}^{2}\left(x^{3}+3x^{2}\right)dx = 0 + 3 \cdot \frac{16}{3} = 16, because the cube is odd and the square is even. Recognising parity turns a two-substitution problem into one line.

Problem. Evaluate \int_{0}^{\pi} x\sin x\,dx two ways.

Solution. By parts. With u = x and dv = \sin x\,dx,

By P4. Replacing x by \pi - x and using \sin(\pi-x) = \sin x,

so 2I = \pi \cdot 2 = 2\pi and I = \pi. The second method needs only \int_{0}^{\pi}\sin x\,dx = 2, and it generalises: for any f, \int_{0}^{\pi}x f(\sin x)\,dx = \frac{\pi}{2}\int_{0}^{\pi}f(\sin x)\,dx.

Problem. Evaluate (a) \int_{1}^{2}\frac{dx}{x(x+1)}, (b) \int_{3}^{5}\frac{dx}{\sqrt{x^{2}-9}}, (c) \int_{0}^{\pi/2}\cos^{3}x\,dx.

Solution. (a) Partial fractions: \frac{1}{x(x+1)} = \frac1x - \frac{1}{x+1}, so

(b) A standard form with a=3:

(c) Write \cos^{3}x = \cos x\left(1-\sin^{2}x\right) and put u = \sin x, with limits u=0 to u=1:

Problem. Evaluate \int_{0}^{1}x^{2}\,dx as the limit of a sum.

Solution. Here a=0, b=1, f(x)=x^{2} and h = \frac1n, so f(a+rh) = \left(\frac{r}{n}\right)^{2}.

Expand the numerator and divide through by n^{3}:

Let n\to\infty, so that \frac3n\to 0 and \frac{1}{n^{2}}\to 0:

This agrees with \left[\frac{x^{3}}{3}\right]_{0}^{1} = \frac13, and the numbers in the figure — 0.4688, 0.3850, 0.3459 — are S_4, S_{10} and S_{40} from this very formula.

Problem. Evaluate \int_{0}^{1}e^{x}\,dx and \int_{0}^{2}\left(x^{2}+1\right)dx as limits of sums.

Solution. (a) With h = \frac1n, f(a+rh) = e^{r/n}, so the sum is geometric with first term e^{1/n} and ratio e^{1/n}:

Write t = \frac1n, so t\to 0 as n\to\infty:

using the standard limit \lim_{t\to 0}\frac{e^{t}-1}{t} = 1. And indeed \left[e^{x}\right]_{0}^{1} = e-1.

(b) Now a=0, b=2, h = \frac2n and f(a+rh) = \frac{4r^{2}}{n^{2}}+1:

which matches \left[\frac{x^{3}}{3}+x\right]_{0}^{2} = \frac83+2 = \frac{14}{3}.

Problem. Find the area bounded by y = x^{2}, the x-axis and the lines x=0 and x=3.

Solution. The curve is above the axis throughout [0,3], so

Similarly \int_{0}^{2}x^{3}dx = \left[\frac{x^{4}}{4}\right]_{0}^{2} = 4 and \int_{1}^{e}\frac{dx}{x} = \left[\ln x\right]_{1}^{e} = 1 — the area under the rectangular hyperbola from 1 to e is exactly one square unit, which is what makes e the natural base.

Problem. Find the area enclosed between the line y = 2x and the parabola y = x^{2}.

Solution. Step 1 — find the intersections. Solve x^{2} = 2x, i.e. x(x-2) = 0, so x = 0 and x = 2.

Step 2 — decide which curve is on top. At x=1, the line gives 2 and the parabola gives 1, so the line is above throughout (0,2).

Step 3 — integrate the difference.

Step 4 — state the answer. The enclosed area is \frac43 square units.

Problem. Find the area bounded by the parabola x = y^{2}, the y-axis and the line y = 2.

Solution. The bounding line is horizontal and the region touches the y-axis, so use horizontal strips and integrate with respect to y. A strip at height y has length x = y^{2} and thickness dy, so

Doing the same region with vertical strips would need \int_{0}^{4}\left(2-\sqrt{x}\right)dx — correct, but longer. Choose the strip direction that makes the limits constants.

Problem. Find the area of the circle x^{2}+y^{2} = a^{2}, and hence the area of the ellipse \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1.

Solution. Circle. By symmetry the total area is four times the first quadrant piece, where y = \sqrt{a^{2}-x^{2}} and x runs from 0 to a:

using Worked example 13.17. At x=a the first term vanishes and \sin^{-1}1 = \frac{\pi}{2}, so

Ellipse. In the first quadrant y = \frac{b}{a}\sqrt{a^{2}-x^{2}}, so the whole calculation is scaled by \frac ba:

With a=4 and b=3 this gives 12\pi, and the shaded quarter in the figure is 3\pi.

Problem. Find the area enclosed between the parabolas y^{2} = 4x and x^{2} = 4y.

Solution. Intersections. From the second, y = \frac{x^{2}}{4}. Substituting into the first, \frac{x^{4}}{16} = 4x, so x^{4} = 64x, giving x\left(x^{3}-64\right) = 0 and hence x = 0 or x = 4.

Upper curve. At x=1, y^{2}=4x gives y = 2 while x^{2}=4y gives y = \frac14. So y = 2\sqrt{x} is above y = \frac{x^{2}}{4}.

Problem. Find the area bounded by y = 4-x^{2} and the x-axis, and the area enclosed between y = x^{2} and y = x^{3}.

Solution. (a) The parabola cuts the axis where 4-x^{2}=0, i.e. x = \pm 2, and is above the axis between them. The function is even, so

(b) x^{2} = x^{3} gives x^{2}(1-x) = 0, so x = 0 and x = 1. On (0,1) we have x^{2} > x^{3} (test x = \frac12: \frac14 > \frac18), so