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Definite Integrals

10.1 FUNDAMENTAL THEOREM OF INTEGRAL CALCULUS

Let \( f \) be a continuous function on the closed interval \( [a, b] \) and \( \phi \) be an anti-derivative of \( f \) , then \[ \begin{array}{l} \int_{a}^{b} f(x) d x=\phi(b)-\phi(a) . \\ \text { (Werds, the above theorem tells us that } \\ \int_{a}^{b} f(x) d x=\text { (value of an anti-derivative at } b, \text { the upper limit) } \\ \qquad \quad-\text { (value of the same anti-derivative at } a, \text { the lower limit). } \end{array} \]

Remarks

1. We often write \( \phi(b)-\phi(a) \) as \( [\phi(x)]_{a}^{b} \) .
2. No matter which anti-derivative we take as \( \phi \) , the value of the definite integral comes out to be the same.

10.1.1 Evaluation of Definite Integrals The fundamental theorem enables us to evaluate the definite integrals by making use of anti- derivatives.

ILLUSTRATIVE EXAMPLES

Example 1. Evaluate the following : (i) \( \int_{0}^{1}\left(2 x^{3}+3\right)^{2} d x \) (ii) \( \int_{3}^{5} \frac{d t}{1+3 t} \)

1. \(\int_0^1\left(2x^3+3\right)^2dx\)
2. \(\int_3^5\frac{dt}{1+3t}\)

Solution. 

1.  \( \int_{0}^{1}\left(2 x^{3}+3\right)^{2} d x=\int_{0}^{1}\left(4 x^{6}+12 x^{3}+9\right) d x \)

    \[ =\left[4 \cdot \frac{x^{7}}{7}+12 \cdot \frac{x^{4}}{4}+9 x\right]_{0}^{1}=\left[\frac{4}{7} x^{7}+3 x^{4}+9 x\right]_{0}^{1} \]

   \[ =\left(\frac{4}{7}+3+9\right)-(0+0+0)=\frac{88}{7} \text {. } \]

2. \[ \begin{aligned} \int_{3}^{5} \frac{d t}{1+3 t} &=\left[\frac{\log |1+3 t|}{3}\right]_{3}^{5}=\frac{1}{3}(\log 16-\log 10) \\ &=\frac{1}{3} \log \frac{16}{10}=\frac{1}{3} \log \frac{8}{5} \end{aligned} \]

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Example 2. Evaluate the following: 

1. \( \int_{0}^{\pi / 2} \cos ^{2} x d x \)
2. \(\int_0^{\frac{\pi}{2}}\cos^4xdx\)

Solution. 

1.  \(\begin{aligned}\int_0^{\pi/2}\cos^2xdx & =\int_0^{\pi/2}\frac{1+\cos2x}{2}dx=\frac{1}{2}[x+ \\ & =\frac{1}{2}\left[\left(\frac{\pi}{2}+\frac{1}{2}\sin\pi\right)-\left(0+\frac{1}{2}\sin\right.\right. \\ & =\frac{1}{2}\left[\left(\frac{\pi}{2}+\frac{1}{2}\cdot0\right)-\frac{1}{2}.0\right]=\frac{\pi}{4}.\end{aligned}\)
2. \( \int_{0}^{\pi / 2} \cos ^{4} x d x=\int_{0}^{\pi / 2}\left(\cos ^{2} x\right)^{2} d x=\int_{0}^{\pi / 2}\left(\frac{1+\cos 2 x}{2}\right)^{2} d x \)

\( =\frac{1}{4} \int_{0}^{\pi / 2}\left(1+2 \cos 2 x+\cos ^{2} 2 x\right) d x=\frac{1}{4} \int_{0}^{\pi / 2}\left(1+2 \cos 2 x+\frac{1+\cos 4 x}{2}\right) d x \)

\( =\frac{1}{8} \int_{0}^{\pi / 2}(3+4 \cos 2 x+\cos 4 x) d x=\frac{1}{8}\left[3 x+4 \cdot \frac{\sin 2 x}{2}+\frac{\sin 4 x}{4}\right]_{0}^{\pi / 2} \)

\( =\frac{1}{8}\left[3\left(\frac{\pi}{2}-0\right)+2(\sin \pi-\sin 0)+\frac{1}{4}(\sin 2 \pi-\sin 0)\right] \)

\[ =\frac{1}{8}\left[\frac{3 \pi}{2}+2(0-0)+\frac{1}{4}(0-0)\right]=\frac{3 \pi}{16} . \]

Example 3. Evaluate the following: 

1.   \( \int_{0}^{\pi / 2} \sqrt{1+\cos 2 x} d x \quad \) 
2.  \( \int_{0}^{\pi / 2} \sqrt{1+\sin 2 x} d x \)

Solution. 

1. \( \int_{0}^{\pi / 2} \sqrt{1+\cos 2 x} d x=\int_{0}^{\pi / 2} \sqrt{2 \cos ^{2} x} d x=\sqrt{2} \int_{0}^{\pi / 2}|\cos x| d x \)

    \(=\sqrt{2}\int_0^{\pi/2}\cos xdx\)

   \( \left(\right. \) As \( 0 \leq x \leq \frac{\pi}{2} \Rightarrow \cos x \geq 0 \Rightarrow|\cos x|=\cos x \) )

   \( =\sqrt{2}[\sin x]_{0}^{\pi / 2}=\sqrt{2}\left[\sin \frac{\pi}{2}-\sin 0\right] \)

   \( =\sqrt{2}(1-0)=\sqrt{2} \) .

2. \( \int_{0}^{\pi / 2} \sqrt{1+\sin 2 x} d x=\int_{0}^{\pi / 2} \sqrt{1+\cos \left(\frac{\pi}{2}-2 x\right)} d x=\int_{0}^{\pi / 2} \sqrt{2 \cos ^{2}\left(\frac{\pi}{4}-x\right)} d x \)

   \( =\sqrt{2} \int_{0}^{\pi / 2}\left|\cos \left(\frac{\pi}{4}-x\right)\right| d x=\sqrt{2} \int_{0}^{\pi / 2} \cos \left(\frac{\pi}{4}-x\right) d x \)

   \( \left[\right. \) As \( 0 \leq x \leq \frac{\pi}{2} \Rightarrow 0 \geq-x \geq-\frac{\pi}{2} \Rightarrow \frac{\pi}{4} \geq \frac{\pi}{4}-x \geq-\frac{\pi}{4} \)

   \[ \left.\Rightarrow-\frac{\pi}{4} \leq \frac{\pi}{4}-x \leq \frac{\pi}{4} \Rightarrow \cos \left(\frac{\pi}{4}-x\right)>0 \Rightarrow\left|\cos \left(\frac{\pi}{4}-x\right)\right|=\cos \left(\frac{\pi}{4}-x\right)\right] \]