Definite Integrals

  • Find the definite integral of algebraic, trigonometric/inverse trigonometric, exponential and logarithmic functions.

  • Use properties of definite integrals.

Properties of Definite Integrals:

  1. Change of variable has no effect on definite integration. \[\int_{a}^{b} f(x)dx = \int_{a}^{b} f(y)dy\]

  2. If limits of definite integration are reversed, then the answer changes sign. (because now the area under the curve given by f(x) is calculated backwards.) \[\int_{a}^{b} f(x)dx = -\int_{b}^{a} f(x)dx\]

    Properties of Definite Integration

  3. Definite integration can be split into two sub-definite integration as long as the point of division is in the interval of original limits of definite integration. \[\int_{a}^{b} f(x)dx = \int_{a}^{c} f(x)dx + \int_{c}^{b} f(x)dx.....a<c<b\]

  4. special cases for odd and even functions.

    Properties of Definite Integration
    For even function, f(-x) = f(x) i.e the graph is symmetrical about y-axis. In such case, \[\int_{-a}^{a} f(x)dx = 2 \int_{0}^{a} f(x)dx\]

    For odd functions, f(-x) = -f(x) and graph of f(x) is mirrored about x-axis AND y-axis. In such case, \[\int_{-a}^{a} f(x)dx = 0\] This is because the ’negative’ area balances the ’positive’ area after the origin.

  5. \[\int_{0}^{2a} f(x)dx = \int_{0}^{a} f(x)dx + \int_{0}^{a} f(2a - x)dx\]

Solved Examples

Solved Example:

2-8-01

\[\int_{-a}^{a}(\sin^6 x + \sin^7 x)dx = ?\]

Solution:
A function is said to be even function if: $f(-x) = f(x)$
A function is said to be odd function if: $f(-x) = -f(x)$
Here, $\sin(x)$ is an odd function because: $\sin(-x) = - \sin x$
$\sin^6 x$, being an even power of $\sin x$ is an even function.
$\sin^7 x$, being an odd power of $\sin x$ is an odd function. \begin{align*} \int_{-a}^{a}(\sin^6 x + \sin^7 x)dx &= \int_{-a}^{a}(\sin^6 x) dx + \int_{-a}^{a}(\sin^7 x)dx\\ &= \int_{-a}^{a}(\sin^6 x) dx + 0\\ &= 2 \int_{0}^{a}(\sin^6 x) dx \end{align*}

Correct Answer: A

Solved Example:

2-8-02

The value of the integral: $\int ^{\infty }_{-\infty }\dfrac {dx}{1+x^{2}}$ is:

Solution:
\begin{align*} I &=\int ^{\infty }_{-\infty }\dfrac {dx}{1+x^{2}}\\ &=\left[ \tan ^{-1}x\right] ^{\infty }_{-\infty }\\ &=\left[ \tan ^{-1}\left(\infty \right) -\tan ^{-1}\left( -\infty \right) \right]\\ &=\dfrac {\pi }{2}-\left( -\dfrac {\pi }{2}\right)\\ &=\pi \end{align*}

Correct Answer: D

Solved Example:

2-8-03

If f (x) is an even function and a is a positive real number, then $\int ^{a}_{-a}f\left( x\right) dx$ equals:

Solution:
For a function, whose limits bounded between -a to a and a is a positive real number. The solution is given by:
If f(x) is an even function, which means if f(-x) = f(x) $\int ^{a}_{-a}f\left( x\right) dx=2\int ^{a}_{0}f\left( x\right) dx$
If f(x) is an odd function, which means if f(-x) = - f(x) $\int ^{a}_{-a}f\left( x\right) dx=0$

Correct Answer: A

Solved Example:

2-8-04

Let f be a continuous and +ve real valued function on [0, 1]. Then $\int_0^\pi f(\sin x) \cos x\ dx = ?$

Solution:
Let t = $\sin x$, So, dt = $\cos x\ dx$
When x = 0, t = 0, When x = $\pi$, t = 0
$\int_0^\pi f(\sin x) \cos x\ dx$ becomes $ \int_0^0 f(t)\ dt$
Since, both upper and lower limits are identical, the value of the integral is zero.

Correct Answer: A

Solved Example:

2-8-05

The value of: \[\int_{-1}^1 x e^{\lvert x \rvert} dx\] is: (GATE Civil 2021)

Solution:

2-8-05

\[f(x) = x e^{\lvert x \rvert}\] \[f(-x) = -x e^{\lvert -x \rvert} = -x e^{\lvert x \rvert} = -f(x)\]

$x e^{\lvert x \rvert}$ is an odd function because f(-x) = -f(x), also the limits are symmetric about zero. Hence, the value of the integral between -1 to 0 cancels out the one from 0 to +1.

Correct Answer: D