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This tutorial covers material encountered in chapter 11 of the VCE Mathematical Methods Textbook, namely:

  • Integration

If you’re thinking “that’s all?” do not be deceived. While derivatives are fairly tame and tedious to derive at worst, non-trivial integrals are normally unable to be evaluated using standard means. Yet many of the most surprising, deep and beautiful results in mathematics are found in the world of integrals. (Un)fortunately for us Maths Methods doesn’t stray far from the most basic of integrals and all of the ones found in the course can be easily evaluated.

Q1 – Evaluate Indefinite Integrals

Q2 – Find Definite Integral from a Given Definite Integral – 1

Q3 – Evaluate Definite Integrals

Q4 – Find Definite Integral from a Given Definite Integral – 2

Q5 – Sketch and Indicate the Area defined by Functions

Q7 – Use a Sum of Definite Integrals to Represent the Shaded Area


Q1. Evaluate the following indefinite integrals:

(a) \int x^4dx

(b) \int \cos(2t)dt

(c) \int (x+2)^3dx

(d) \int \dfrac{dx}{3x}

(e) \int e^{tx}dx

Q2. Find \int_{-2}^{2}(x+3f(x))dx if \int_{-2}^{2} f(x)dx=6

Q3. Evaluate each of the following definite integrals:

(a) \int_{0}^{2\pi}\sin(2x)dx

(b) \int_{-5}^{5}(2e^x+2e^{-x})dx

(c) \int_{-5}^{5}(2e^x-2e^{-x})dx

(d) \int_{1}^{e}(\dfrac{1}{x}+x^2)dx

(e) \int_{-3}^{3}(x^2-x)dx

Q4. If \int_{0}^{2}f(x)dx=5 find \int_{\frac{1}{2}}^{\frac{3}{2}}f(2x-1)

Q5. (a) Sketch f(x)=e^x and g(x)=e^{-x} on one set of axis and clearly indicate, by shading the region, the area given by \int_{0}^{3}f(x)dx+\int_{-3}^{0}g(x)dx

(b) evaluate \int_{0}^{3}f(x)dx + \int_{-3}^{0}g(x)dx

Q6. Sketch a graph of h(x)=5e^{\frac{x}{3}-3} and find the area of the region enclosed between the curve, the y-axis and the line x=5

Q7. Set up a sum of definite integrals that represents the shaded area between the curves y=f(x) and y=g(x) as shown below:

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